problem solving with kinematic equations

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Kinematics Practice Problems with Answers

The most complete guide on solving kinematics problems for high schools and colleges. By working through these questions, you can undoubtedly master this topic in physics. 

These multiple-choice questions on kinematics for AP Physics 1 are also available to review for students enrolled in AP Physics courses.

All kinematics equations are summarized in the following expressions: \begin{gather*} \Delta x=\frac{v_1+v_2}{2}\times \Delta t \\\\ v=v_0+at \\\\ \Delta x=\frac 12 at^2+v_0t \\\\ v^2-v_0^2=2a\Delta x \end{gather*} In the rest of this long article, you will see how to apply these equations in the given problems.

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Kinematics Practice Problems:

Problem (1): A car slows down its motion from 10 m/s to 6 m/s in 2 seconds under constant acceleration.  (a) What is its acceleration? (b) How far did the car travel during this time interval?

Solution : This is a basic kinematics problem, so we will explain the steps in detail.

Step 1: Since all these problems are in one dimension, draw a horizontal axis (like the positive $x$ axis), and place the object on it, so that its motion matches the direction of the axis. 

Step 2:  Specify the known and wanted information. Here, in the elapsed time interval $2\,{\rm s}$, the initial and final velocities of the car are given as $v_i=10\,{\rm m/s}$ and $v_f=6\,{\rm m/s}$. The wanted quantity is the constant acceleration of the object (car), $a=?$. 

Step 3: Apply the kinematics equation that is appropriate for this situation. 

(a) To find the acceleration in this problem, we are given the time, initial, and final velocity. The kinematics equation $v=v_0+at$ is suitable for this situation, as the only unknown variable is the acceleration $a$. By rearranging the equation, we get \begin{gather*} v=v_0+at\\\\ 6=10+a(2) \\\\ 6-10=2a \\\\\Rightarrow \quad a=\frac{6-10}{2}=-2\quad {\rm \frac{m}{s^2}}\end{gather*} Since the problem states that the acceleration is constant, we could also use any of the other constant acceleration kinematics equations. The negative sign of the acceleration indicates that it is directed toward the negative $x$ axis.

(b)  "How far'' means the distance traveled by car is wanted, denoted by $x$ in the kinematics equations. 

Here, the best equation that relates the known and unknown information is $x=\frac 12 at^2+v_0t$ or $v^2-v_0^2=2ax$. We choose the first, so \begin{align*} x&=\frac 12 at^2+v_0t \\\\&=\frac 12 (-2)(2)^2+(10)(2) \\\\&=16\quad {\rm m}\end{align*} 

On the following page you can find over 40+ questions related to applying kinematics equations in velocity and acceleration:

Velocity and acceleration problems

Problem (2): A moving object slows down from $12\,{\rm m/s}$ to rest at a distance of 20 m. Find the acceleration of the object (assumed constant).

Solution : In the diagram below, all known information along with the direction of the uniform motion is shown. 

A common phrase in kinematics problems is “ending or coming to a rest”, which means the final velocity of the object in the time interval we are considering is zero, $v_f=0$.

The kinematics equation that suits this problem is $v^2-v_0^2=2a(x-x_0)$, where the only unknown variable is the acceleration $a$. 

For simplicity, we can assume the initial position of the motion $x_0$ is zero in all kinematics problems, $x_0$. \begin{gather*} v^2-v_0^2=2ax\\\\0^2-(12)^2 =2a(20) \\\\ \rightarrow a=\frac{-144}{2\times 20}\\\\\Rightarrow \boxed{a=-3.6\quad {\rm \frac{m}{s^2}}}\end{gather*} As before, the negative sign indicates of the acceleration indicates that it is directed to the left .

Problem (3): A bullet leaves the muzzle of an 84-cm rifle with a speed of 521 m/s. Find the magnitude of the bullet's acceleration by assuming it is constant inside the barrel of the rifle.

Solution : The bullet accelerates from rest to a speed of 521 m/s over a distance of 0.84 meters. We have the following known quantities: the initial velocity $v_0$, ​ the final velocity $v$, and the displacement $x-x_0$. The unknown is acceleration $a$. The perfect kinematics equation that relates all these variables is $v^2-v_0^2=2a(x-x_0)$. Solving for $a$, we have: \begin{gather*}v^2-v_0^2=2a(x-x_0)\\\\ (521)^2-0=2(a)(0.84-0) \\\\ \Rightarrow \boxed{a=1.62\times 10^5\quad {\rm m/s^2}}\end{gather*} This results in a very large acceleration. 

Problem (4): A car starts its motion from rest and uniformly accelerates at a rate of $4\,{\rm m/s^2}$ for 2 seconds in a straight line.  (a) How far did the car travel during those 2 seconds?  (b) What is the car's velocity at the end of that time interval?

Solution : "Start from rest'' means the initial object's velocity is zero, $v_0=0$. The known information are $a=4\,{\rm m/s^2}$, $t=2\,{\rm s}$ and wants the distance traveled $x=?$. 

(a) The kinematics equation that relates that information is $x=\frac 12 at^2+v_0 t+x_0$ since the only unknown quantity is $x$ with the given known data above. \begin{align*} x&=\frac 12 at^2+v_0 t+x_0 \\\\&=\frac 12 (4)(2)^2+(0)(2) \\\\&=\boxed{8\quad {\rm m}}\end{align*} As previous, we set $x_0=0$. 

(b) Now that the distance traveled by the car in that time interval is known, we can use the following kinematics equation to find the car's final velocity $v$. \begin{align*} v^2-v_0^2 &=2a(x-x_0) \\\\v^2-(0)^2&=2(4)(8-0) \\\\v^2&=64\end{align*} Taking the square root, we get $v$: \[v=\sqrt{64}=\pm 8\quad {\rm \frac ms}\] We know that velocity is a vector quantity in physics and has both a direction and a magnitude. 

The magnitude of the velocity (speed) was obtained as 8 m/s, but in what direction? Or we must choose which signs? Because the car is uniformly accelerating without stopping in the positive $x$ axis, the correct sign for velocity is positive. 

Therefore, the car's final velocity is $\boxed{v_f=+8\,{\rm m/s}}$. 

Challenging Kinematics Problems

In the following, some challenging kinematics problems are presented, which are for homework.

A driver is moving along at $45\,\rm m/s$ when she suddenly notices a roadblock $100\,\rm m$ ahead. Can the driver stop the vehicle in time to avoid colliding with the obstruction if her reaction time is assumed to be $0.5\,\rm s$ and her car's maximum deceleration is $5\,\rm m/s^2$?

Solution : The time between seeing the obstacle and taking action, such as slamming on the brake, is defined as the reaction time. During this time interval, the moving object travels at a constant speed. 

Thus, in all such questions, we have two phases. One is constant speed, and the other is accelerating with negative acceleration (deceleration). 

Here, between the time of seeing the barrier and the time of braking, the driver covers a distance of \begin{align*} x_1&=vt_{reac} \\\\ &=25\times 0.5 \\\\ &=12.5\,\rm m\end{align*} In the decelerating phase, the car moves a distance, which is obtained using the following kinematics equation: \begin{gather*} v^2-v_0^2=2ax_2 \\\\ (0)^2-(25)^2=2(-5) x_2 \\\\ \Rightarrow x_2=62.5\,\rm m\end{gather*} Summing these two distances gives a total distance that is a good indication of whether the moving object hits the obstacle or not. \begin{gather*} \Delta x_{actual}=x_1+x_2=75\,\rm m \\\\ \Rightarrow \Delta x_{covered}<\Delta x_{actual} \end{gather*} As a result, because the distance covered by the car is less than the actual distance between the time of seeing the barrier and the obstacle itself, the driver has sufficient time to stop the car in time to avoid a collision.

For a moving car at a constant speed of $90\,\rm km/h$ and a human reaction time of $0.3\,\rm s$; find the stopping distance if it slows down at a rate of $a=3\,\rm m/s^2$. 

Solution : We use SI units, so first convert the given speed in these units as below \begin{align*} v&=90\,\rm km/h \\\\ &=\rm 90\times \left(\frac{1000\,m}{3600\,s}\right) \\\\ &=25\,\rm m/s\end{align*} As we said previously, during the reaction time, your car moves at a constant speed and covers a distance of \begin{align*} x_1&=vt_{react} \\\\ &=(25)(0.3) \\\\ &=7.5\,\rm m \end{align*} Deceleration means the moving object slows down, or a decrease per second in the velocity of the car occurs. In this case, we must put the acceleration with a negative sign in the kinematics equations. 

During the second phase, your car has negative acceleration and wants to be stopped. Thus, the distance covered during this time interval is found as follows \begin{align*} v^2-v_0^2=2a\Delta x \\\\ (0)^2-(25)^2=2(3)\Delta x \\\\ \Rightarrow \quad \boxed{\Delta x=104.17\,\rm m}\end{align*} 

Assume you are traveling $35\,\rm m/s$ when suddenly you see red light traffic $50\,\rm m$ ahead. If it takes you $0.456\,\rm s$ to apply the brakes and the maximum deceleration of the car is $4.5\,\rm m/s^2$,  (a) Will you be able to stop the car in time?  (b) How far from the time of seeing the red light will you be? 

Solution : When you see the red light until you apply the brakes, your car is moving at a constant speed. This time interval is defined as the reaction time, $\Delta t_{react}=0.456\,\rm s$. After you get the brakes on, the car starts to decelerate at a constant rate, $a=-4.5\,\rm m/s^2$. Pay attention to the negative signs of such problems. The negative tells us that the car is decreasing its speed. 

(a) In the first phase, the car moves a distance of \begin{align*} x_1&=v\Delta t_{react} \\\\ &=35\times 0.455 \\\\ &=15.96\,\rm m\end{align*} In the phase of deceleration, the car is moving a distance whose magnitude is found using the time-independent kinematics equation as below \begin{gather*} v^2-v_0^2=2ax_2 \\\\ (0)^2-(35)^2=2(4.5)x_2 \\\\ \Rightarrow \quad x_2=136.11\,\rm m\end{gather*} The sum of these two distances traveled gives us the total distance covered by the car from the time of seeing the red traffic light to the moment of a complete stop. \[x_{tot}=x_1+x_2=152.07\,\rm m \] Because the total distance traveled is greater than the actual distance to the red light, the driver will not be able to stop the car in time. 

(b) As previously calculated, the total distance traveled by the car is nearly $152\,\rm m$ or the car is about $102\,\rm m$ past the red light traffic. 

A person stands on the edge of a $60-\,\rm m$-high cliff and throws two stones vertically downward, $1$ second apart, and sees they both reach the water simultaneously. The first stone had an initial speed of $4\,\rm m/s$.  (a) How long after dropping the first stone does the second stone hit the water? (b) How fast was the second stone released? (c) What is the speed of each stone at the instant of hitting the water?

Solution:  Because all quantities appearing in the kinematics equation are vectors, we must first choose a positive direction. Here, we take up as a positive $y$ direction.

Both stones arrived in the water at the same time. Thus, calculate the time the first stone was in the air. Next,  use the time interval between the two drops to find the duration the second stone was in the air. (a) The first stone is released downward at a speed of $4\,\rm m/s$, thus, its initial velocity is $v_0=-4\,\rm m/s$. The minus sign is for moving in the opposite direction of the chosen direction. 

The only relevant kinematics equation that relates this known information is $\Delta y=-\frac 12 gt^2+v_0t$, where $\Delta y=-60\,\rm m$ is the vertical displacement, and the negative indicates that the stone hit a point below the chosen origin. Substituting the numerical values into this and solving for the time duration $t$ gives \begin{gather*} \Delta y=-\frac 12 gt^2+v_0t \\\\ -60=-\frac 12 (10)t^2+(-4)t \\\\ 5t^2+4t-60=0 \\\\ \Rightarrow \boxed{t=3.0\,\rm s} \, , \, t'=-3.8\,\rm s \end{gather*} The second answer is not acceptable. 

(b) The second stone was released $1$ second after throwing the first one and arrived at the same time as the first stone. Therefore, the time interval that the second stone was in the air is found to be \begin{align*} t_2&=t_1-1 \\ &=3.0-1\\ &=2\,\rm s\end{align*} 

(c) It is better to apply the time-independent kinematics equation $v^2-v_0^2=-2g\Delta y$ to find the stone's velocity at the moment it hit the water. For the first stone, we have \begin{gather*} v^2-v_0^2=-2g\Delta y \\\\ v^2-(-4)^2=-2(10)(-60) \\\\ \Rightarrow \quad \boxed{v=34.8\,\rm m/s} \end{gather*} The second stone's velocity is left to you as an exercise.

In a tennis game, the ball leaves the racket at a speed of $75\,\rm m/s$ whereas it is in contact with the racket for $25\,\rm ms$, and starts at rest. Assume the ball experiences constant acceleration.  What was the ball's acceleration during this serve? How far has the ball traveled on this serve?

Solution : In this question, we are asked to find the ball's acceleration and distance traveled during that pretty small time interval.  (a) In a time interval of $\Delta t=25\times 10^{-3}\,\rm s$, we are given the beginning velocity $v_1=0$ and the end velocity $v_2=85\,\rm m/s$. Because it is assumed the acceleration is constant, the average acceleration definition, $a=\frac{\Delta v}{\Delta t}$, is best suited for these known quantities. \begin{align*} a&=\frac{v_2-v_1}{\Delta t} \\\\ &=\frac{75-0}{25\times 10^{-3}} \\\\ &=3000\,\rm m/s^2 \end{align*} A huge acceleration is given to the tennis ball.  (b) Here, we are asked to find the amount of distance traveled by the ball during the time the ball was in contact with the racket. Because we have a constant acceleration motion, it is best to use the following equation to find the distance traveled. \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\times \Delta t \\\\ &=\frac{0+75}{2}\times (25\times 10^{-3}) \\\\ &=937.5\times 10^{-3}\,\rm m\end{align*} In millimeters, $\Delta x=937.5\,\rm mm$, and in centimeters $\Delta x=93.75\,\rm cm$. Therefore, during this incredibly short time interval, the ball moves about $94\,\rm cm$ along with the racket.

Starting from rest and ending at rest, a car travels a distance of $1500\,\rm m$ along the $x$-axis. During the first quarter of the distance, it accelerates at a rate of $+1.75\,\rm m/s^2$, while for the remaining distance, its acceleration is $-0.450\,\rm m/s^2$.  (a) What is the time travel of the whole path?  (b) What is the maximum speed of the car over this distance?

None of the time-dependent kinematics equations give us the time travel $t'$ without knowing the initial speed at the instant of the start of this second path. 

We can find it using the equation $v^2-v_0^2=2a\Delta x$, setting $v=0$ at the end of the path, and solving for $v_0$ \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ (0)^2-v_0^2=2(-0.450)(1125) \\\\ v_0=\sqrt{2\times 0.45\times 1125} \\\\ \Rightarrow v_0=31.82\,\rm m/s\end{gather*} Given that, one can use the simple equation $v=v_0+at$ and solve for the time travel in this part of the path. \begin{gather*} v=v_0+at \\\\ 0=31.82+(-0.450)t' \\\\ \Rightarrow t'=70.71\,\rm s\end{gather*} Therefore, the total time traveled over the entire path is the sum of these two times. \begin{align*} T&=t+t' \\\\ &=20.70+70.71 \\\\ &=\boxed{91.41\,\rm s} \end{align*}

A train that is $75$ meters long starts accelerating uniformly from rest. When the front of the train reaches a railway worker who is standing $150$ meters away from where the train started, it is traveling at a speed of $20\,\rm m/s$. What will be the speed of the last car as it passes the worker?

Solution : The front of the train is initially $150\,\rm m$ away from the worker, and when it passes him, it has a speed of $25\,\rm m/s$. From this data, we can find the acceleration of the front of the train (which is the same acceleration as the whole train) by applying the following kinematics equation \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ (25)^2-(0)^2=2a\times 150 \\\\ \Rightarrow \quad a=2.08\,\rm m/s^2\end{gather*} Given the train's acceleration, now focus on the last car.

The last car is initially at rest and placed at a distance of $150+80=230\,\rm m$ away from the person. When it passes the person, it has traveled $\Delta x= 230\,\rm m$ and its speed is determined simply as below \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ v^2-(0)^2=2(2.08)(230) \\\\ \Rightarrow \quad \boxed{v=30.93\,\rm m/s}\end{gather*}

A wildcat moving with constant acceleration covers a distance of $100\,\rm m$ apart in $8\,\rm s$. Assuming that its speed at the second point is $20\,\rm m/s$,  (a) What was its speed in the first place?  (b) At what rate does its speed change over this distance? 

Solution : First of all, list all known data given to us. Time interval $\Delta t=8\,\rm s$, the horizontal displacement $\Delta x=100\,\rm m$, speed at second point $v_2=20\,\rm m/s$. 

We are asked to find the speed at the second point. To solve this kinematics problem, we use the following kinematics equation because the acceleration is constant and this is the most relevant equation that relates the known to the unknown quantities. \begin{gather*} \Delta x=\frac{v_1+v_2}{2}\times \Delta t \\\\ 100=\frac{v_1+20}{2}\times 8 \\\\ \Rightarrow \quad \boxed{v_1=5\,\rm m/s} \end{gather*} Therefore, the wildcat's speed in the first place is $5\,\rm m/s$. 

In this part, we should find the wildcat's acceleration because acceleration is defined as the time rate of change of the speed of a moving object. Given the first place speed, $v_1=5\,\rm m/s$, found in the preceding part, we can use the following time-independent kinematics equation to find the wanted unknown. \begin{gather*} v_2^2-v_1^2=2a\Delta x \\\\ (20)^2-(5)^2=2a(100) \\\\ \Rightarrow  \quad \boxed{a=1.875\,\rm m/s^2}\end{gather*}

A car slows down uniformly from $45\,\rm m/s$ to rest in $10\,\rm s$. How far did it travel in this time interval?

Solution : List the data known as follows: initial speed $v_0=45\,\rm m/s$, final speed $v=0$, and the total time duration that this happened is $t=10\,\rm s$. The unknown is also the amount of displacement, $\Delta x$. 

The only kinematics equation that relates those together is $\Delta x=\frac{v_1+v_2}{2}\times \Delta t$, where $v_1$ and $v_2$ are the velocities at the beginning and end of that time interval. \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\times \Delta t \\\\ &=\frac{45+0}{2}\times 10 \\\\ &=\boxed{225\,\rm m}\end{align*} Keep in mind that we use this formula when the object slows down uniformly, or, in other words when the object's acceleration is constant.

Problem (5): We want to design an airport runway with the following specifications. The lowest acceleration of a plane should be $4\,{\rm m/s^2}$ and its take-off speed is 75 m/s. How long would the runway have to be to allow the planes to accelerate through it? 

Solution: The known quantities are $a=4\,{\rm m/s^2}$, and final velocity $v=75\,{\rm m/s}$. The wanted quantity is runway length $\Delta x=x-x_0$. The perfect kinematics equation that relates those together is $v^2-v_0^2=2a(x-x_0)$. \begin{align*} v^2-v_0^2&=2a\Delta x\\\\ (75)^2-0&=2(4) \Delta x\\\\ \Rightarrow \Delta x&=\boxed{703\quad {\rm m}}\end{align*} Thus, if the runway wants to be effective, its length must be at least about 703 meters.

Problem (6): A stone is dropped vertically from a high cliff. After 3.55 seconds, it hits the ground. How high is the cliff? 

Solution : There is another type of kinematics problem in one dimension but in the vertical direction. In such problems, the constant acceleration is that of free falling, $a=g=-10\,{\rm m/s^2}$.  

"Dropped'' or "released'' in free-falling problems means the initial velocity is zero, $v_0=0$.  In addition, it is always better to consider the point of release as the origin of the coordinate, so $y_0=0$. 

The most relevant kinematics equation for these known and wanted quantities is $y=-\frac 12 gt^2+v_0t+y_0$ \begin{align*} y&=-\frac 12 gt^2+v_0t+y_0 \\\\&=-\frac 12 (9.8)(3.55)^2+0+0\\\\&=\boxed{-61.8\quad {\rm m}}\end{align*} The negative indicates that the impact point is below our chosen origin . 

Problem (7):  A ball is thrown into the air vertically from the ground level with an initial speed of 20 m/s.  (a) How long is the ball in the air? (b) At what height does the ball reach?

Solution : The throwing point is considered to be the origin of our coordinate system, so $y_0=0$. Given the initial velocity $v_0=+20\,{\rm m/s}$ and the gravitational acceleration $a=g=-9.8\,{\rm m/s^2}$. The wanted time is how long it takes the ball to reach the ground again.

To solve this free-fall problem , it is necessary to know some notes about free-falling objects. 

Note (1): Because the air resistance is neglected, the time the ball is going up is half the time it is going down.  

Note (2): At the highest point of the path, the velocity of the object is zero. 

(a) By applying the kinematics equation $v=v_0+at$ between the initial and the highest ($v=0$) points of the vertical path, we can find the going up time. \begin{align*} v&=v_0+at \\0&=20+(-9.8)t\\\Rightarrow t&=2.04\quad {\rm s}\end{align*} The total flight time is twice this time \[t_{tot}=2t=2(2.04)=4.1\,{\rm s}\] Hence, the ball takes about 4 seconds to reach the ground. 

(b) The kinematics equation $v^2-v_0^2=2a(y-y_0)$ is best for this part. \begin{align*} v^2-v_0^2&=2a(y-y_0) \\0-20^2&=2(-9.8)(y-0) \\ \Rightarrow y&=\boxed{20\quad {\rm m}}\end{align*} Hence, the ball goes up to a height of about 20 meters.

Problem (8): An object moving in a straight line with constant acceleration, has a velocity of $v=+10\,{\rm m/s}$ when it is at position $x=+6\,{\rm m}$ and of $v=+15\,{\rm m/s}$ when it is at $x=10\,{\rm m}$. Find the acceleration of the object.

Solution : Draw a diagram, put all known data into it, and find a relevant kinematics equation that relates them together. 

We want to analyze the motion in a distance interval of $\Delta x=x_2-x_1=10-6=4\,{\rm m}$, thus, we can consider the velocity at position $x_1=6\,{\rm m}$ as the initial velocity and at $x_2=10\,{\rm m}$ as the final velocity. 

The most relevant kinematics equation that relates these known quantities to the wanted acceleration $a$ is $v^2-v_0^2=2a(x-x_0)$, where $x-x_0$ is the same given distance interval. Thus, \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\ (15)^2-(10)^2&=2(a)(4) \\\\225-100&=8a\\\\\Rightarrow a&=\frac{125}{8}\\\\&=15.6\,{\rm m/s^2}\end{align*} 

Problem (9): A moving object accelerates uniformly from 75 m/s at time $t=0$ to 135 m/s at $t=10\,{\rm s}$. How far did it move at the time interval $t=2\,{\rm s}$ to $t=4\,{\rm s}$? 

Solution : Draw a diagram and implement all known data in it as below. 

Because the problem tells us that the object accelerates uniformly, its acceleration is constant along the entire path. 

Given the initial and final velocities of the moving object, its acceleration is determined using the definition of instantaneous acceleration as below \[a=\frac{v_2-v_1}{t_2-t_1}=\frac{135-75}{10}=6\,{\rm m/s^2}\] In this kinematics problem, to analyze the motion between the requested times (stage II in the figure), we must have a little bit of information for that time interval, their velocities, or the distance between them. 

As you can see in the figure, the initial velocity of stage II is the final velocity of stage I. By using a relevant kinematics equation that relates those data to each other, we would have \begin{align*} v&=v_0+at\\\\&=75+(6)(2) \\\\&=87\,{\rm m/s}\end{align*} This velocity would be the initial velocity for stage II of the motion. Now, all known information for stage II is initial velocity $v_0=87\,{\rm m/s}$, acceleration $a=6\,{\rm m/s^2}$, and time interval $\Delta t=2\,{\rm s}$. The wanted is the distance traveled $x=?$

The appropriate equation that relates all these together is $x=\frac 12 at^2+v_0t+x_0$. \begin{align*}x&=\frac 12 at^2+v_0t+x_0\\\\&=\frac 12 (6)(2)^2+(87)(2)+0\\\\&=186\quad {\rm m}\end{align*} Hence, our moving object travels a distance of 186 m between the instances of 2 s and 4 s. 

Problem (10): From rest, a fast car accelerates with a uniform rate of $1.5\,{\rm m/s^2}$ in 4 seconds. After a while, the driver applies the brakes for 3 seconds, causing the car to uniformly slow down at a rate of $-2\,{\rm m/s^2}$.  (a) How fast is the car at the end of the braking period? (b) How far has the car traveled after braking?

Solution : This motion is divided into two parts. First, draw a diagram and specify each section's known kinematics quantities. 

(a) In the first part, given the acceleration, initial velocity, and time interval, we can find its final velocity at the end of 4 seconds. \begin{align*} v&=v_0+at\\&=0+(1.5)(4) \\&=6\quad {\rm m/s}\end{align*} This velocity is considered as the initial velocity for the second part, whose final velocity is wanted. 

In the next part, the acceleration magnitude and braking time interval are given, so its final velocity is found as below \begin{align*} v&=v_0+at\\&=6+(-2)(3) \\&=0\end{align*} The zero velocity here indicates that the car, after the braking period, comes to a stop. 

(b) The distance traveled in the second part is now calculated using the kinematics equation $x=\frac 12 at^2+v_0t+x_0$, because the only unknown quantity is distance $x$. \begin{align*} x&=\frac 12 at^2+v_0t+x_0\\\\&=\frac 12 (-2)(3)^2+(6)(3)+0\\\\&=+9\quad {\rm m}\end{align*} Therefore, after braking, the car traveled a distance of 9 meters before getting stopped. 

Problem (11): A car moves at a speed of 20 m/s down a straight path. Suddenly, the driver sees an obstacle in front of him and applies the brakes. Before the car reaches a stop, it experiences an acceleration of $-10\,{\rm m/s^2}$.  (a) After applying the brakes, how far did it travel before stopping?  (b) How long does it take the car to reach a stop? 

Solution : As always, the first and most important step in solving a kinematics problem is drawing a diagram and putting all known values into it, as shown below.

(a) The kinematics equation $v^2-v_0^2=2a(x-x_0)$ is the perfect equation as the only unknown quantity in it is the distance traveled $x$. Thus, \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\ 0^2-(20)^2&=2(-10)(x-0) \\\\\Rightarrow \quad x&=\frac {-400}{-20}\\\\&=20\quad {\rm m}\end{align*} (b) "how long does it take'' asks us to find the time interval. The initial and final velocities, as well as acceleration, are known, so the only relevant kinematics equation is $v=v_0+at$. Thus, \begin{align*} v&=v_0+at\\\\0&=20+(-10)t\\\\\Rightarrow t&=\frac{-20}{-10}\\\\&=2\quad {\rm s}\end{align*} Therefore, after braking, the car has moved for 2 seconds before reaching a stop.

Problem (12): A sports car moves a distance of 100 m in 5 seconds with a uniform speed. Then, the driver brakes, and the car, come to a stop after 4 seconds. Find the magnitude and direction of its acceleration (assumed constant). 

Solution : uniform speed means constant speed or zero acceleration for the motion before braking. Thus, we can use the definition of average velocity to find its speed just before braking as below \begin{align*} \bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{100}{5}\\\\&=20\quad {\rm m/s}\end{align*} Now, we know the initial and final velocities of the car in the braking stage. Since the acceleration is assumed to be constant, by applying the definition of average acceleration, we would have \begin{align*} \bar{a}&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-20}{4}\\\\&=-5\quad {\rm m/s^2}\end{align*} The negative shows the direction of the acceleration, which is toward the negative $x$-axis. 

Hence, the car's acceleration has a magnitude of $5\,{\rm m/s^2}$ in the negative $x$ direction. 

Problem (13): A race car accelerates from rest at a constant rate of $2\,{\rm m/s^2}$ in 15 seconds. It then travels at a constant speed for 20 seconds, and after that, it comes to a stop with an acceleration of $2\,{\rm m/s^2}$.  (a) What is the total distance traveled by car? (b) What is its average velocity over the entire path?

Solution : To solve this kinematics question, we divided the entire path into three parts. 

Part I: "From rest'' means the initial velocity is zero. Thus, given the acceleration and time interval, we can use the kinematics equation $v=v_0+at$ to calculate the distance traveled by car at the end of 15 seconds for the first part of the path. \begin{align*} x&=\frac{1}{2}at^2 +v_0 t+x_0\\\\&=\frac 12 (2)(15)^2 +(0)(15)+0\\\\&=\boxed{125\quad{\rm m}}\end{align*}   As a side calculation, we find the final velocity for this part as below \begin{align*}v&=v_0+at\\\\&=0+(2) (15) \\\\&=30\quad {\rm m/s}\end{align*} Part II: the speed in this part is the final speed in the first part because the car continues moving at this constant speed after that moment. 

The constant speed means we are facing zero acceleration. As a result, it is preferable to use the average velocity definition rather than the kinematics equations for constant (uniform) acceleration.

The distance traveled for this part, which takes 20 seconds at a constant speed of 30 m/s, is computed by the definition of average velocity as below \begin{align*} \bar{v}&=\frac{\Delta x}{\Delta t}\\\\30&=\frac{\Delta x}{20}\end{align*} Thus, we find the distance traveled as $\boxed{x=600\,{\rm m}}$. 

Part III: In this part, the car comes to a stop, $v=0$, so its acceleration must be a negative value as $a=-2\,{\rm m/s^2}$. Here, the final velocity is also zero. Its initial velocity is the same as in the previous part. 

Consequently, the best kinematics equation that relates those known to the wanted distance traveled $x$, is $v^2-v_0^2=2a(x-x_0)$. \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\0^2-(30)^2&=2(-2)(x-0) \\\\ \Rightarrow \quad x&=\boxed{225\quad {\rm m}}\end{align*} The total distance traveled by car for the entire path is the sum of the above distances \[D=125+600+225=\boxed{950\quad {\rm m}}\]

Problem (14): A ball is dropped vertically downward from a tall building of 30-m-height with an initial speed of 8 m/s. After what time interval does the ball strike the ground? (take $g=-10\,{\rm m/s^2}$.)

Solution : This is a free-falling kinematics problem. As always, choose a coordinate system along with the motion and the origin as the starting point. 

Here, the dropping point is considered the origin, so in all kinematics equations, we set $y_0=0$. By this choice, the striking point is 30 meters below the origin, so in equations, we also set $y=-30\,{\rm m}$. 

Remember that velocity is a vector in physics whose magnitude is called speed. In this problem, the initial speed is 8 m/s downward. This means that the velocity vector is written as $v=-8\,{\rm m/s}$. 

Now that all necessary quantities are ready, we can use the kinematics equation $y=\frac 12 at^2+v_0t+y_0$, to find the wanted time that the ball strikes the ground. \begin{align*} y&=\frac 12 at^2+v_0t+y_0\\\\-30&=\frac 12 (-10)t^2+(-8)t+0\end{align*} After rearranging, a quadratic equation like $5t^2+8t-30=0$ is obtained, whose solutions are given as below: \begin{gather*} t=\frac{-8\pm\sqrt{8^2-4(5)(-30)}}{2(5)}\\\\ \boxed{t_1=1.77\,{\rm s}} \quad , \quad t_2=-3.37\,{\rm s}\end{gather*} $t_1$ is the accepted time because the other is negative, which is not acceptable in kinematics. Therefore, the ball takes about 1.7 seconds to hit the ground. 

Note: The solutions of a quadratic equation like $at^2+bt+c=0$, where $a,b,c$ are some constants, are found by the following formula: \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] 

Problem (15): The acceleration versus time graph for an object that moves at a constant speed of 30 m/s is shown in the figure below. Find the object's average velocity between instances $t_1=10\,{\rm s}$ and $t_2=30\,{\rm s}$. 

Solution : The best and shortest approach to solving such a kinematics problem is to first draw its velocity-vs-time graph. Next, the area under the obtained graph gives us the total displacement, which is divided by the total time interval to yield the average velocity.

The path consists of three parts with different accelerations. 

In the first part, the object slows down its motion at a constant rate of $-2\,{\rm m/s^2}$ in 10 seconds. Its initial velocity is also 30 m/s. With these known quantities in hand, the kinematics equation $v=v_0+at$ gives us the velocity at the end of this time interval. \begin{align*} v&=v_0+at\\&=30+(-2)(10) \\&=10\quad {\rm m/s}\end{align*} This calculation corresponds to a straight line between the points $(v=30\,{\rm m/s},t=0)$ and $(v=10\,{\rm m/s},t=10\,{\rm s})$ on the $v-t$ graph as shown below.

Next, the object moves with zero acceleration for 5 seconds, which means the velocity does not change during this time interval. This implies that we must draw a horizontal line in the $v-t$ graph.

In the last part, the object accelerates from 10 m/s with a constant rate of $+2\,{\rm m/s^2}$ in 15 seconds. Thus, its final velocity at the end of this time interval is determined as below \begin{align*} v&=v_0+at\\&=10+(2)(15) \\&=40\quad {\rm m/s}\end{align*} Now, it's time to draw the velocity-vs-time graph. As an important point, note that all these motions have a constant acceleration, so all parts of a velocity-time graph, are composed of straight-line segments with different slopes.

For part I, we must draw a straight-line segment between the velocities of 30 m/s and 10 m/s. 

Part II is a horizontal line since its velocities are constant during that time interval, and finally, in Part III, there is a straight line between velocities of 10 m/s and 40 m/s. 

All these verbal phrases are illustrated in the following velocity-vs-time graph . 

Recall that the area under a velocity vs. time graph always gives the displacement. Hence, the area under the $v-t$ graph between 10 s and 30 s gives the displacement. Therefore, the areas of rectangle $S_1$ and trapezoid $S_2$ are calculated as below \begin{gather*} S_1 =10\times 5=50\quad {\rm m} \\\\S_2=\frac{10+40}{2}\times 15=375\quad {\rm m}\end{gather*} Therefore, the total displacement in the time interval $[15,30]$ is \[D=S_{tot}=S_1+S_2=425\,{\rm m}\] From the definition of average velocity, we have \[\bar{v}=\frac{displacement}{time}=\frac{425}{20}=21.25\,{\rm m/s}\] 

In this tutorial, all concepts about kinematics equations are taught in a problem-solution strategy. All these answered problems are helpful for MCAT physics exams. 

We can also find these kinematic variables using a position-time  or velocity-time graph. Because slopes in those graphs represent velocity and acceleration, respectively, and the concavity of a curve in a position vs. time graph shows the sign of its acceleration in an x-t graph  as well.

Author : Dr. Ali Nemati Date published : 8-7-2021 Updated : June 12, 2023

© 2015 All rights reserved. by Physexams.com

Kinematic Equations: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: April 29, 2022

Now that you’ve learned about displacement, velocity, and acceleration, you’re well on your way to being able to describe just about any motion you could observe around you with physics. All that’s left is to learn how these values really play into each other. We know a few ways to move between them, but they’re all pretty limited. What happens if you need to find displacement, but only know acceleration and time? We don’t have a way to combine all of those values yet. Enter the four kinematic equations. 

What We Review

The Kinematic Equations

The following four kinematic equations come up throughout physics from the earliest high school class to the highest level college course:

Don’t let all of these numbers and symbols intimidate you. We’ll talk through each one – what they mean and when we use them. By the end of this post, you’ll be a master of understanding and implementing each of these physics equations. Let’s start with defining what all of those symbols mean. 

The First Kinematic Equation

This physics equation would be read as “the final velocity is equal to the initial velocity plus acceleration times time”. All it means is that if you have constant acceleration for some amount of time, you can find the final velocity. You’ll use this one whenever you’re looking at changing velocities with a constant acceleration.

The Second Kinematic Equation

This one is read as “displacement equals final velocity plus initial velocity divided by two times time”. You’ll use this one whenever you don’t have an acceleration to work with but you need to relate a changing velocity to a displacement.

The Third Kinematic Equation

This one may look a bit scarier as it is longer than the others, but it is read as “displacement equals initial velocity times time plus one half acceleration times time squared”. All it means is that our displacement can be related to our initial velocity and a constant acceleration without having to find the final velocity. You’ll use this one when final velocity is the only value you don’t know yet.

It is worth noting that this kinematic equation has another popular form: x=x_{0}+v_{0}t+\frac{1}{2}at^{2} . While that may seem even more intimidating, it’s actually exactly the same. The only difference here is that we have split up \Delta x into x-x_{0} and then solved to get x on its own. This version can be particularly helpful if you’re looking specifically for a final or initial position rather than just an overall displacement.

The Fourth Kinematic Equation

Our last kinematic equation is read as “final velocity squared equals initial velocity squared plus two times acceleration times displacement”. It’s worth noting that this is the only kinematic equation without time in it. Many starting physicists have been stumped by reaching a problem without a value for time. While staring at an equation sheet riddled with letters and numbers can be overwhelming, remembering you have this one equation without time will come up again and again throughout your physics career.

It may be worth noting that all of these are kinematic equations for constant acceleration. While this may seem like a limitation, we learned before that high school physics courses generally utilize constant acceleration so we don’t need to worry about it changing yet. If you do find yourself in a more advanced course, new physics equations will be introduced at the appropriate times.

How to Approach a Kinematics Problem

So now that we have all of these different kinematic equations, how do we know when to use them? How can we look at a physics word problem and know which of these equations to apply? You must use problem-solving steps. Follow these few steps when trying to solve any complex problems, and you won’t have a problem.

Step 1: Identify What You Know

This one probably seems obvious, but skipping it can be disastrous to any problem-solving endeavor. In physics problems, this just means pulling out values and directions. If you can add the symbol to go with the value (writing t=5\text{ s} instead of just 5\text{ s} , for example), even better. It’ll save time and make future steps even easier.

Step 2: Identify the Goal

In physics, this means figuring out what question you’re actually being asked. Does the question want you to find the displacement? The acceleration? How long did the movement take? Figure out what you’re being asked to do and then write down the symbol of the value you’re solving for with a question mark next to it ( t=\text{?} , for example). Again, this feels obvious, but it’s also a vital step.

Step 3: Gather Your Tools

Generally, this means a calculator and an equation. You’ll want to look at all of the symbols you wrote down and pick the physics equation for all of them, including the unknown value. Writing everything down beforehand will make it easier to pull a relevant equation than having to remember what values you need while searching for the right equation. You can use the latter method, but you’re far more likely to make a mistake and feel frustrated that way.

Step 4: Put it all Together

Plug your values into your equation and solve for the unknown value. This will usually be your last step, though you may find yourself having to repeat it a few times for exceptionally complex problems. That probably won’t come up for quite a while, though. After you’ve found your answer, it’s generally a good idea to circle it to make it obvious. That way, whoever is grading you can find it easily and you can easily keep track of which problems you’ve already completed while flipping through your work.

Kinematic Equation 1: Review and Examples

To learn how to solve problems with these new, longer equations, we’ll start with v=v_{0}+at . This kinematic equation shows a relationship between final velocity, initial velocity, constant acceleration, and time. We will explore this equation as it relates to physics word problems. This equation is set up to solve for velocity, but it can be rearranged to solve for any of the values it contains. For this physics equation and the ones following, we will look at one example finding the variable that has already been isolated and one where a new variable needs to be isolated using the steps we just outlined. So, let’s jump into applying this kinematic equation to a real-world problem.

A car sits at rest waiting to merge onto a highway. When they have a chance, they accelerate at 4\text{ m/s}^2 for 7\text{ s} . What is the car’s final velocity?

We have a clearly stated acceleration and time, but there’s no clearly defined initial velocity here. Instead, we have to take this from context. We know that the car “sits at rest” before it starts moving. This means that our initial velocity in this situation is zero. Other context clues for an object starting at rest is if it is “dropped” or if it “falls”. Our other known values will be even easier to pull as we were actually given numerical values. Now it’s time to put everything into a list.

• v_{0}=0\text{ m/s}
• a=4\text{ m/s}^2
• t=7\text{ s}

Our goal here was clearly stated: find the final velocity. We’ll still want to list that out so we can see exactly what symbols we have to work with on this problem.

We already know which of the kinematic equations we’re using, but if we didn’t, this would be where we search our equation sheet for the right one. Regardless, we’ll want to write that down too.

Step 4: Put it All Together

At this point, we’ll plug all of our values into our kinematic equation. If you’re working on paper, there’s no need to repeat anything we’ve put above. That being said, for the purposes of digital organization and so you can see the full problem in one spot, we will be rewriting things here.

Now let’s get a bit trickier with a problem that will require us to rearrange our kinematic equation.

A ball rolls toward a hill at 3\text{ m/s} . It rolls down the hill for 5\text{ s} and has a final velocity of 18\text{ m/s} . What was the ball’s acceleration as it rolled down the hill?

Just like before, we’ll make a list of our known values:

• v_{0}=3\text{ m/s}
• t=5\text{ s}
• v=18\text{ m/s}

Again, our goal was clearly stated, so let’s add it to our list:

We already know which equation we’re using, but let’s pretend we didn’t. We know that we need to solve for acceleration, but if you look at our original list of kinematic equations, there isn’t one that’s set up to solve for acceleration:

This begs the question, how to find acceleration (or any value) that hasn’t already been solved for? The answer is to rearrange an equation. First, though, we need to pick the right one. We start by getting rid of the second equation in this list as it doesn’t contain acceleration at all. Our options are now:

• \Delta x=v_{0}t+\dfrac{1}{2}at^{2}
• v^{2}=v_{0}^{2}+2a\Delta x

Now we’ll need to look at the first list we made of what we know. We know the initial velocity, time, and final velocity. There’s only one equation that has all the values we’re looking for and all of the values we know with none that we don’t. This is the first kinematic equation:

In this case, we knew the kinematic equation coming in so this process of elimination wasn’t necessary, but that won’t often be the case in the future. You’ll likely have to find the correct equation far more often than you’ll have it handed to you. It’s best to practice finding it now while we only have a few equations to work with.

Like before, we’ll be rewriting all of our relevant information below, but you won’t need to if you’re working on paper.

Although you can plug in values before rearranging the equation, in physics, you’ll usually see the equation be rearranged before values are added. This is mainly done to help keep units where they’re supposed to be and to avoid any mistakes that could come from moving numbers and units rather than just a variable. We’ll be taking the latter approach here. Follow the standard PEMDAS rules for rearranging the equation and then write it with the variable we’ve isolated on the left. While that last part isn’t necessary, it is a helpful organizational practice:

For a review of solving literal equations, visit this post ! Now we can plug in those known values and solve:

Kinematic Equation 2: Review and Examples

Next up in our four kinematics equations is \Delta x=\dfrac{v+v_{0}}{2} t . This one relates an object’s displacement to its average velocity and time. The right-hand side shows the final velocity plus the initial velocity divided by two – the sum of some values divided by the number of values, or the average. Although this equation doesn’t directly show a constant acceleration, it still assumes it. Applying this equation when acceleration isn’t constant can result in some error so best not to apply it if a changing acceleration is mentioned.

A car starts out moving at 10\text{ m/s} and accelerates to a velocity of 24\text{ m/s} . What displacement does the car cover during this velocity change if it occurs over 10\text{ s} ?

• v_{0}=10\text{ m/s}
• v=24\text{ m/s}
• t=10\text{ s}
• \Delta x=\text{?}
• \Delta x=\dfrac{v+v_{0}}{2} t

This time around we won’t repeat everything here. Instead, We’ll jump straight into plugging in our values and solving our problem:

A ball slows down from 15\text{ m/s} to 3\text{ m/s} over a distance of 36\text{ m} . How long did this take?

• v_{0}=15\text{ m/s}
• v=3\text{ m/s}
• \Delta x=36\text{ m}

We don’t have a kinematic equation for time specifically, but we learned before that we can rearrange certain equations to solve for different variables. So, we’ll pull the equation that has all of the values we need and isolate the variable we want later:

Again, we won’t be rewriting anything, but we will begin by rearranging our equation to solve for time:

Now we can plug in our known values and solve for time.

Kinematic Equation 3: Review and Examples

Our next kinematic equation is \Delta x=v_{0}t+\frac{1}{2}at^{2} . This time we are relating our displacement to our initial velocity, time, and acceleration. The only odd thing you may notice is that it doesn’t include our final velocity, only the initial. This equation will come in handy when you don’t have a final velocity that was stated either directly as a number or by a phrase indicating the object came to rest. Just like before, we’ll use this equation first to find a displacement, and then we’ll rearrange it to find a different value.

A rocket is cruising through space with a velocity of 50\text{ m/s} and burns some fuel to create a constant acceleration of 10\text{ m/s}^2 . How far will it have traveled after 5\text{ s} ?

• v_{0}=50\text{ m/s}
• a=10\text{ m/s}^2
• \Delta x=v_{0}t+\frac{1}{2}at^{2}

At this point, it appears that these problems seem to be quite long and take several steps. While that is an inherent part of physics in many ways, it will start to seem simpler as time goes on. This problem presents the perfect example. While it may have been easy to combine lines 4 and 5 mathematically, they were shown separately here to make sure the process was as clear as possible. While you should always show all of the major steps of your problem-solving process, you may find that you are able to combine some of the smaller steps after some time of working with these kinematic equations.

Later in its journey, the rocket is moving along at 20\text{ m/s} when it has to fire its thrusters again. This time it covers a distance of 500\text{ m} in 10\text{ s} . What was the rocket’s acceleration during this thruster burn?

• v_{0}=20\text{ m/s}
• \Delta x=500\text{ m}

As usual, we’ll begin by rearranging the equation, this time to solve for acceleration.

Now we can plug in our known values to find the value of our acceleration.

Kinematic Equation 4: Review and Examples

The last of the kinematic equations that we will look at is v^{2}=v_{0}^{2}+2a\Delta x . This one is generally the most complicated looking, but it’s also incredibly important as it is our only kinematic equation that does not involve time. It relates final velocity, initial velocity, acceleration, and displacement without needing a time over which a given motion occurred. For this equation, as with the others, let’s solve it as is and then rearrange it to solve for a different variable.

A car exiting the highway begins with a speed of 25\text{ m/s} and travels down a 100\text{ m} long exit ramp with a deceleration (negative acceleration) of 3\text{ m/s}^2 . What is the car’s velocity at the end of the exit ramp?

• v_{0}=25\text{ m/s}
• \Delta x=100\text{ m}
• a=-3\text{ m/s}^2

Note that our acceleration here is a negative value. That is because our problem statement gave us a deceleration instead of an acceleration. Whenever you have a deceleration, you’ll make the value negative to use it as an acceleration in your problem-solving. This also tells us that our final velocity should be less than our initial velocity so we can add that to the list of what we know as well.

• Final velocity will be less than initial.

Being able to know something to help check your answer at the end is what makes this subject a bit easier than mathematics for some students.

While we generally try to not have any operations going on for the isolated variable, sometimes it’s actually easier that way. Having your isolated variable raised to a power is generally a time to solve before simplifying. This may seem like an arbitrary rule, and in some ways it is, but as you continue through your physics journey you’ll come up with your own practices for making problem-solving easier.

Now that we have both sides simplified, we’ll take the square root to eliminate the exponent on the left-hand side:

If we remember back at the beginning, we said that our final velocity would have to be less than our initial velocity because the problem statement told us that we were decelerating. Our initial velocity was 25\text{ m/s} which is, indeed, greater than 5\text{ m/s} so our answer checks out.

A ghost is sliding a wrench across a table to terrify the mortal onlooker. The wrench starts with a velocity of 2\text{ m/s} and accelerates to a velocity of 5\text{ m/s} over a distance of 7\text{ m} . What acceleration did the ghost move the wrench with?

• v_{0}=2\text{ m/s}
• v=5\text{ m/s}
• \Delta x=7\text{ m}

We can also make an inference about our acceleration here – that it will be positive. Not every problem will tell you clearly the direction of the acceleration, but if your final velocity is greater than your initial velocity, you can be sure that your acceleration will be positive.

• Positive acceleration

You’ll get better at picking up on subtle hints like this as you continue your physics journey and your brain starts naturally picking up on some patterns. You’ll likely find this skill more and more helpful as it develops and as problems get more difficult.

We’ll start by rearranging our equation to solve for acceleration.

As usual, now that we’ve rearranged our equation, we can plug in our values.

Again, we can go back to the beginning when we said our acceleration would be a positive number and confirm that it is. 

Problem-Solving Strategies

At this point, you’re likely getting the sense that physics will be a lot of complex problem-solving. If so, your senses are correct. In many ways, physics is the science of explaining nature with mathematical equations. There’s a lot that goes into developing and applying these equations, but at this point in your physics career, you’ll find that the majority of your time will likely be spent on applying equations to word problems. If you feel that your problem-solving skills could still use some honing, check out more examples and strategies from this post by the Physics Classroom or through this video-guided tutorial from Khan Academy.

That was a lot of equations and examples to take in. Eventually, whether you’re figuring out how to find a constant acceleration or how to solve velocity when you don’t have a value for time, you’ll know exactly which of the four kinematic equations to apply and how. Just keep the problem-solving steps we’ve used here in mind, and you’ll be able to get through your physics course without any unsolvable problems.

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Kinematic Equations

Learn about the kinematic equations., kinematic equations lesson, the four kinematic equations.

• Equation 1:      v = v 0 + at
• Equation 2:      v 2 = v 0 2 + 2a(Δx)
• Equation 3:      x = x 0 + v 0 t
• Equation 4:      x = x 0 + v 0 t + 1 / 2 at 2

Kinematic Variables

• x - Displacement
• v - Velocity
• a - Acceleration

These are the four variables at play with the kinematic equations. The equations describe the motion of an object that is subject to constant acceleration. By leveraging the equations, we can solve for the initial and final values of these variables.

Important: Directionality of the Variables

Displacement, velocity, and acceleration are all directional, whereas time is non-directional. Directional variables will have a positive value when their vector points in the positive direction, and a negative value when their vector points in the negative direction. Since time is non-directional, it will always have a positive value.

When using the kinematic equations to solve problems, it is helpful to choose a direction for positive x , v , and a . Sometimes this direction is given in the problem statement. The opposite of the positive direction will be the negative direction.

When to use the Kinematic Equations

Kinematics is the study of object motion without reference to the forces that cause motion. The kinematic equations are simplifications of object motion. Three of the equations assume constant acceleration (equations 1, 2, and 4), and the other equation assumes zero acceleration and constant velocity (equation 3).

When an object motion problem falls into these categories, we may use the kinematic equations to solve it. For example, we can use them to figure out how far a projectile flies as long as the projectile experiences constant acceleration during its flight. Another example of a simple and effective use of the kinematic equations is when a car driving at constant velocity. Since it has zero acceleration, we may use the equation that has no acceleration terms (equation 3).

How to Choose a Kinematic Equation

We must decide which kinematic equation is best for what we are solving. Generally, kinematics problems involve solving for some unknown. The unknown could be initial displacement, final displacement, change in displacement, initial velocity, final velocity, acceleration, or time.

We choose an equation based on what is known and what is not known. In some cases, we must use several equations sequentially to find the value of our unknown. The example problems below give more insight for the process of choosing an equation.

Kinematic Equations Example Problems

Kinematic equation 1 example.

A projectile is fired from the chamber of a cannon and accelerates at 1500 m/s 2 for 0.75 seconds before leaving the barrel. What is the projectile’s velocity as it leaves the barrel of the cannon?

• Since the projectile is not moving before it is fired, our initial velocity will be zero. We are given the values of acceleration (1500 m/s 2 ) and time (0.75 seconds), so the equation we will use is v = v 0 + at .
• Since we know the values of all variables except final velocity, we may plug in our known values to find v, this gives us v = (0 m/s) + (1500 m/s 2 )(0.75 seconds) = 1125 m/s .
• The projectile's velocity is 1125 m/s as it leaves the barrel.

Kinematic Equation 2 Example

A sprinter is moving at 5 m/s when they reach 20 meters into their race. They maintain a constant acceleration of 2 m/s 2 through 40 meters into the race. What is the sprinter’s velocity 40 meters into the race?

• We know the values of initial velocity (5 m/s), acceleration (2 m/s 2 ), and change in displacement (40 – 20 = 20 meters). The equation we will use is v 2 = v 0 2 + 2a(Δx) .
• Since we know the values of all variables except final velocity, we may plug in our known values and find v. v 2 = (5 m/s) 2 + 2(2 m/s 2 )(20 meters) v 2 = 105 v = 10.25 m/s
• The sprinter's velocity is 10.25 m/s at 40 meters into the race.

Kinematic Equation 3 Example

A car is 200 meters away from a building and starts driving away from the building at 20 m/s. How far away is the car from the building after driving for 6 seconds?

• We know the values of initial displacement (200 meters), initial velocity (20 m/s), and time in motion (6 seconds). We must find final displacement. The kinematic equation we will use is x = x 0 + v 0 t .
• Since we know the values of all variables but one, we may plug in our known values to find the unknown value of x. x = (200 meters) + (20 m/s)(6 seconds) = 320 meters
• After driving for 6 seconds, the car is 320 meters away from the building.

Kinematic Equation 4 Example

A person is standing 6 meters behind you. They throw a ball over your head with a horizontal velocity of 20 m/s. The ball experiences a constant 4 m/s 2 of horizontal deceleration while in flight and takes 2 seconds to land. How far in front of you does the ball land?

• We know the values of initial displacement (-6 meters), initial velocity (20 m/s), acceleration (-4 m/s 2 ), and time (2 seconds). The equation we will use is x = x 0 + v 0 t + 1 / 2 at 2 .
• Since we know the values of all variables except final displacement, we may plug in our known values and find x. x = (-6 meters) + (20 m/s)(2 seconds) + 1 / 2 (-4 m/s 2 )(2 seconds) 2 x = -6 + 40 - 8 x = 26 meters
• The ball lands 26 meters in front of you.

Bonus Kinematics Lesson

What is projectile motion.

Earlier we mentioned that projectile motion problems can be solved by using the kinematic equations. While projectile motion is just one of the many types of problems we come across in kinematics, it is very useful to understand exactly what it is.

By definition, projectile motion is the motion of an object or particle that is only affected by the force of gravity. A projectile follows a parabolic trajectory similar to when a ball is thrown. This parabolic trajectory is also called a ballistic trajectory.

How to Solve Projectile Motion

Since we neglect the force of air resistance in projectile motion, we can figure out how far a projectile will fly by decomposing its velocity into vertical and horizontal components. The horizontal component of velocity will be constant during the flight. The vertical component of velocity will affect the projectile's time of flight, or hang time.

Here's the equations for a projectile's horizontal and vertical motion. They are derived from the kinematic equations.

• v x = v 0 cos(α)
• x = v 0 cos(α)t + x 0
• v y = v 0 sin(α)
• t = 2v y / g

Projectile Motion Example

A ball is thrown from the surface of Earth at 25 m/s on an angle of 30° above the horizontal. Neglecting air resistance, how far does the ball travel horizontally before hitting the ground?

• First, let's set our known variables. v 0 = 25 m/s α = 30° g = 9.81 m/s 2 .
• Let's find the velocity components. v x = v 0 cos(α) = 25cos(30°) = 21.65 m/s v y = v 0 sin(α) = 25sin(30°) = 12.5 m/s
• Now we can find the hang time, and use the hang time to calculate horizontal distance traveled. t = 2v y / g = 2(12.5) / 9.81 = 2.55 x = v 0 cos(α)t + x 0 = 25cos(30°)(2.55) + 0 = (21.65)(2.55) = 55.21
• The ball travels horizontally for 55.21 meters before hitting the ground.

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Chapter 2: Vectors and Scalars

Chapter 3: motion along a straight line, chapter 4: motion in two or three dimensions, chapter 5: newton's laws of motion, chapter 6: application of newton's laws of motion, chapter 7: work and kinetic energy, chapter 8: potential energy and energy conservation, chapter 9: linear momentum, impulse and collisions, chapter 10: rotation and rigid bodies, chapter 11: dynamics of rotational motions, chapter 12: equilibrium and elasticity, chapter 13: fluid mechanics, chapter 14: gravitation, chapter 15: oscillations, chapter 16: waves, chapter 17: sound, chapter 18: temperature and heat, chapter 19: the kinetic theory of gases, chapter 20: the first law of thermodynamics, chapter 21: the second law of thermodynamics, chapter 22: electric charges and fields, chapter 23: gauss's law, chapter 24: electric potential, chapter 25: capacitance, chapter 26: current and resistance, chapter 27: direct-current circuits, chapter 28: magnetic forces and fields, chapter 29: sources of magnetic fields, chapter 30: electromagnetic induction, chapter 31: inductance, chapter 32: alternating-current circuits, chapter 33: electromagnetic waves.

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The kinematic equations of motion are useful to solve problems involving one-dimensional motion of objects under constant acceleration.

Consider a couple driving to a nearby coffee shop. They start the car and apply a constant acceleration of 2 meters per second squared. What will be the car's velocity after 20 seconds and the distance covered by it in that time?

The choice of the equation to solve the problem depends on the known quantities and the unknown quantities.

Here, the known quantities are constant acceleration, time, the initial position, and the initial velocity, as the car was at rest. The unknown quantities are the velocity and distance covered after 20 seconds, which can be calculated using the first and second kinematic equations.

Substituting known values in the first kinematic equation gives the velocity of the car, which equals 40 meters per second. 

Then, substituting the known values in the second kinematic equation, simplifying and solving it gives the distance covered by the car equal to 400 meters.

3.10: Kinematic Equations: Problem Solving

When analyzing one-dimensional motion with constant acceleration, the problem-solving strategy involves identifying the known quantities and choosing the appropriate kinematic equations to solve for the unknowns. Either one or two kinematic equations are needed to solve for the unknowns, depending on the known and unknown quantities. Generally, the number of equations required is the same as the number of unknown quantities in the given example. Two-body pursuit problems always require two equations to be solved simultaneously to derive the value of the unknown.

In complex problems, it is not always possible to identify the unknowns or the order in which they should be calculated. In such scenarios, it is useful to make a list of unknowns and draw a sketch of the problem to identify the directions of motion of an object. To solve the problem, substitute the knowns along with their units into the appropriate equation. This step provides a numerical answer, and also provides a check on units that can help find errors. If the units are incorrect, then an error has been made. However, correct units do not necessarily guarantee that the numerical part of the answer is also correct.

The final step in solving problems is to check the answer to see if it is reasonable. This final step is crucial as the goal of physics is to describe nature accurately. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. This enables us to get a conceptual understanding of the problems that are being solved. Sometimes the physical principle may be applied correctly to solve the numerical problem, but produces an unreasonable result. For example, if an athlete starting a foot race accelerates at 0.4 m/s² for 100 seconds, their final speed will be 40 m/s (about 150 km/h). This result is unreasonable because a person cannot run at such a high speed for 100 seconds. Here, the physics is correct in a sense, but there is more to describing nature than just manipulating equations correctly.

This text is adapted from Openstax, University Physics Volume 1, Section 3.4: Motion with Constant Acceleration .

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• 5.3 Projectile Motion
• Introduction
• 1.1 Physics: Definitions and Applications
• 1.2 The Scientific Methods
• 1.3 The Language of Physics: Physical Quantities and Units
• Section Summary
• Key Equations
• Concept Items
• Critical Thinking Items
• Performance Task
• Multiple Choice
• Short Answer
• Extended Response
• 2.1 Relative Motion, Distance, and Displacement
• 2.2 Speed and Velocity
• 2.3 Position vs. Time Graphs
• 2.4 Velocity vs. Time Graphs
• 3.1 Acceleration
• 3.2 Representing Acceleration with Equations and Graphs
• 4.2 Newton's First Law of Motion: Inertia
• 4.3 Newton's Second Law of Motion
• 4.4 Newton's Third Law of Motion
• 5.1 Vector Addition and Subtraction: Graphical Methods
• 5.2 Vector Addition and Subtraction: Analytical Methods
• 5.4 Inclined Planes
• 5.5 Simple Harmonic Motion
• 6.1 Angle of Rotation and Angular Velocity
• 6.2 Uniform Circular Motion
• 6.3 Rotational Motion
• 7.1 Kepler's Laws of Planetary Motion
• 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity
• 8.1 Linear Momentum, Force, and Impulse
• 8.2 Conservation of Momentum
• 8.3 Elastic and Inelastic Collisions
• 9.1 Work, Power, and the Work–Energy Theorem
• 9.2 Mechanical Energy and Conservation of Energy
• 9.3 Simple Machines
• 10.1 Postulates of Special Relativity
• 10.2 Consequences of Special Relativity
• 11.1 Temperature and Thermal Energy
• 11.2 Heat, Specific Heat, and Heat Transfer
• 11.3 Phase Change and Latent Heat
• 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium
• 12.2 First law of Thermodynamics: Thermal Energy and Work
• 12.3 Second Law of Thermodynamics: Entropy
• 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators
• 13.1 Types of Waves
• 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period
• 13.3 Wave Interaction: Superposition and Interference
• 14.1 Speed of Sound, Frequency, and Wavelength
• 14.2 Sound Intensity and Sound Level
• 14.3 Doppler Effect and Sonic Booms
• 14.4 Sound Interference and Resonance
• 15.1 The Electromagnetic Spectrum
• 15.2 The Behavior of Electromagnetic Radiation
• 16.1 Reflection
• 16.2 Refraction
• 16.3 Lenses
• 17.1 Understanding Diffraction and Interference
• 17.2 Applications of Diffraction, Interference, and Coherence
• 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge
• 18.2 Coulomb's law
• 18.3 Electric Field
• 18.4 Electric Potential
• 18.5 Capacitors and Dielectrics
• 19.1 Ohm's law
• 19.2 Series Circuits
• 19.3 Parallel Circuits
• 19.4 Electric Power
• 20.1 Magnetic Fields, Field Lines, and Force
• 20.2 Motors, Generators, and Transformers
• 20.3 Electromagnetic Induction
• 21.1 Planck and Quantum Nature of Light
• 21.2 Einstein and the Photoelectric Effect
• 21.3 The Dual Nature of Light
• 22.1 The Structure of the Atom
• 22.2 Nuclear Forces and Radioactivity
• 22.3 Half Life and Radiometric Dating
• 22.4 Nuclear Fission and Fusion
• 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation
• 23.1 The Four Fundamental Forces
• 23.2 Quarks
• 23.3 The Unification of Forces
• A | Reference Tables

Section Learning Objectives

By the end of this section, you will be able to do the following:

• Describe the properties of projectile motion
• Apply kinematic equations and vectors to solve problems involving projectile motion

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (C) analyze and describe accelerated motion in two dimensions using equations.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Motion in Two Dimensions, as well as the following standards:

• (C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.

Section Key Terms

Properties of projectile motion.

Projectile motion is the motion of an object thrown (projected) into the air when, after the initial force that launches the object, air resistance is negligible and the only other force that object experiences is the force of gravity. The object is called a projectile , and its path is called its trajectory . Air resistance is a frictional force that slows its motion and can significantly alter the trajectory of the motion. Due to the difficulty in calculation, only situations in which the deviation from projectile motion is negligible and air resistance can be ignored are considered in introductory physics. That approximation is often quite accurate.

[BL] [OL] Review addition of vectors graphically and analytically.

[BL] [OL] [AL] Explain the term projectile motion. Ask students to guess what the motion of a projectile might depend on? Is the initial velocity important? Is the angle important? How will these things affect its height and the distance it covers? Introduce the concept of air resistance. Review kinematic equations.

The most important concept in projectile motion is that when air resistance is ignored, horizontal and vertical motions are independent , meaning that they don’t influence one another. Figure 5.27 compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly.

Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical.

We’ll call the horizontal axis the x -axis and the vertical axis the y -axis. For notation, d is the total displacement, and x and y are its components along the horizontal and vertical axes. The magnitudes of these vectors are x and y , as illustrated in Figure 5.28 .

As usual, we use velocity, acceleration, and displacement to describe motion. We must also find the components of these variables along the x - and y -axes. The components of acceleration are then very simple a y = – g = –9.80 m/s 2 . Note that this definition defines the upwards direction as positive. Because gravity is vertical, a x = 0. Both accelerations are constant, so we can use the kinematic equations. For review, the kinematic equations from a previous chapter are summarized in Table 5.1 .

Where x is position, x 0 is initial position, v is velocity, v avg is average velocity, t is time and a is acceleration.

Solve Problems Involving Projectile Motion

The following steps are used to analyze projectile motion:

• Separate the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ A x = A cos θ and A y = A sin θ A y = A sin θ are used. The magnitudes of the displacement s s along x- and y-axes are called x x and y . y . The magnitudes of the components of the velocity v v are v x = v ​ ​ ​ cos θ v x = v ​ ​ ​ cos θ and v y = v ​ ​ ​ sin θ v y = v ​ ​ ​ sin θ , where v v is the magnitude of the velocity and θ θ is its direction. Initial values are denoted with a subscript 0.
• Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms Horizontal Motion ( a x = 0 ) x = x 0 + v x t v x = v 0 x = v x = velocity  is a constant. Horizontal Motion ( a x = 0 ) x = x 0 + v x t v x = v 0 x = v x = velocity  is a constant. Vertical motion (assuming positive is up a y = − g = − 9.80  m/s 2 a y = − g = − 9.80  m/s 2 ) y = y 0 + 1 2 ( v 0 y + v y ) t v y = v 0 y − g t y = y 0 + v 0 y t − 1 2 g t 2 v y 2 = v 0 y 2 − 2 g ( y − y 0 ) y = y 0 + 1 2 ( v 0 y + v y ) t v y = v 0 y − g t y = y 0 + v 0 y t − 1 2 g t 2 v y 2 = v 0 y 2 − 2 g ( y − y 0 )
• Solve for the unknowns in the two separate motions (one horizontal and one vertical). Note that the only common variable between the motions is time t t . The problem solving procedures here are the same as for one-dimensional kinematics.

Teacher Demonstration

Demonstrate the path of a projectile by doing a simple demonstration. Toss a dark beanbag in front of a white board so that students can get a good look at the projectile path. Vary the toss angles, so different paths can be displayed. This demonstration could be extended by using digital photography. Draw a reference grid on the whiteboard, then toss the bag at different angles while taking a video. Replay this in slow motion to observe and compare the altitudes and trajectories.

Tips For Success

For problems of projectile motion, it is important to set up a coordinate system. The first step is to choose an initial position for x x and y y . Usually, it is simplest to set the initial position of the object so that x 0 = 0 x 0 = 0 and y 0 = 0 y 0 = 0 .

Watch Physics

Projectile at an angle.

This video presents an example of finding the displacement (or range) of a projectile launched at an angle. It also reviews basic trigonometry for finding the sine, cosine and tangent of an angle.

• The time to reach the ground would remain the same since the vertical component is unchanged.
• The time to reach the ground would remain the same since the vertical component of the velocity also gets doubled.
• The time to reach the ground would be halved since the horizontal component of the velocity is doubled.
• The time to reach the ground would be doubled since the horizontal component of the velocity is doubled.

Worked Example

A fireworks projectile explodes high and away.

During a fireworks display like the one illustrated in Figure 5.30 , a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75° above the horizontal. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

The motion can be broken into horizontal and vertical motions in which a x = 0 a x = 0 and   a y = g   a y = g . We can then define x 0 x 0 and y 0 y 0 to be zero and solve for the maximum height .

By height we mean the altitude or vertical position y y above the starting point. The highest point in any trajectory, the maximum height, is reached when   v y = 0   v y = 0 ; this is the moment when the vertical velocity switches from positive (upwards) to negative (downwards). Since we know the initial velocity, initial position, and the value of v y when the firework reaches its maximum height, we use the following equation to find y y

Because y 0 y 0 and v y v y are both zero, the equation simplifies to

Solving for y y gives

Now we must find v 0 y v 0 y , the component of the initial velocity in the y -direction. It is given by v 0 y = v 0 sin θ v 0 y = v 0 sin θ , where v 0 y v 0 y is the initial velocity of 70.0 m/s, and θ = 75 ∘ θ = 75 ∘ is the initial angle. Thus,

Since up is positive, the initial velocity and maximum height are positive, but the acceleration due to gravity is negative. The maximum height depends only on the vertical component of the initial velocity. The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding.

There is more than one way to solve for the time to the highest point. In this case, the easiest method is to use y = y 0 + 1 2 ( v 0 y + v y ) t y = y 0 + 1 2 ( v 0 y + v y ) t . Because y 0 y 0 is zero, this equation reduces to

Note that the final vertical velocity, v y v y , at the highest point is zero. Therefore,

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + v 0 y t − 1 2 g t 2 y = y 0 + v 0 y t − 1 2 g t 2 , and solving the quadratic equation for t t .

Because air resistance is negligible, a x = 0 a x = 0 and the horizontal velocity is constant. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t x = x 0 + v x t , where x 0 x 0 is equal to zero

where v x v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0 . v x = v 0 cos θ 0 . Now,

The time t t for both motions is the same, and so x x is

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below, while some of the fragments may now have a velocity in the –x direction due to the forces of the explosion.

[BL] [OL] [AL] Talk about the sample problem. Discuss the variables or unknowns in each part of the problem Ask students which kinematic equations may be best suited to solve the different parts of the problem.

The expression we found for y y while solving part (a) of the previous problem works for any projectile motion problem where air resistance is negligible. Call the maximum height y = h y = h ; then,

This equation defines the maximum height of a projectile . The maximum height depends only on the vertical component of the initial velocity.

Calculating Projectile Motion: Hot Rock Projectile

Suppose a large rock is ejected from a volcano, as illustrated in Figure 5.31 , with a speed of 25.0   m / s 25.0   m / s and at an angle 3 5 ° 3 5 ° above the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path.

Breaking this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the time. The time a projectile is in the air depends only on its vertical motion.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

If we take the initial position y 0 y 0 to be zero, then the final position is y = − 20.0  m . y = − 20.0  m . Now the initial vertical velocity is the vertical component of the initial velocity, found from

Substituting known values yields

Rearranging terms gives a quadratic equation in t t

This expression is a quadratic equation of the form a t 2 + b t + c = 0 a t 2 + b t + c = 0 , where the constants are a = 4.90, b = –14.3, and c = –20.0. Its solutions are given by the quadratic formula

This equation yields two solutions t = 3.96 and t = –1.03. You may verify these solutions as an exercise. The time is t = 3.96 s or –1.03 s. The negative value of time implies an event before the start of motion, so we discard it. Therefore,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m / s 14.3 m / s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

Practice Problems

The fact that vertical and horizontal motions are independent of each other lets us predict the range of a projectile. The range is the horizontal distance R traveled by a projectile on level ground, as illustrated in Figure 5.32 . Throughout history, people have been interested in finding the range of projectiles for practical purposes, such as aiming cannons.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 v 0 , the greater the range, as shown in the figure above. The initial angle θ 0 θ 0 also has a dramatic effect on the range. When air resistance is negligible, the range R R of a projectile on level ground is

where v 0 v 0 is the initial speed and θ 0 θ 0 is the initial angle relative to the horizontal. It is important to note that the range doesn’t apply to problems where the initial and final y position are different, or to cases where the object is launched perfectly horizontally.

Virtual Physics

Projectile motion.

In this simulation you will learn about projectile motion by blasting objects out of a cannon. You can choose between objects such as a tank shell, a golf ball or even a Buick. Experiment with changing the angle, initial speed, and mass, and adding in air resistance. Make a game out of this simulation by trying to hit the target.

Check Your Understanding

• Projectile motion is the motion of an object projected into the air and moving under the influence of gravity.
• Projectile motion is the motion of an object projected into the air and moving independently of gravity.
• Projectile motion is the motion of an object projected vertically upward into the air and moving under the influence of gravity.
• Projectile motion is the motion of an object projected horizontally into the air and moving independently of gravity.

What is the force experienced by a projectile after the initial force that launched it into the air in the absence of air resistance?

• The nuclear force
• The gravitational force
• The electromagnetic force
• The contact force

Use the Check Your Understanding questions to assess whether students achieve the learning objectives for this section. If students are struggling with a specific objective, the Check Your Understanding will help identify which objective is causing the problem and direct students to the relevant content.

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Kinematic Equations

Sensors & datalogging, curriculum & bundles, lab apparatus & supplies, stem sense solutions, product guides.

What are the Kinematic Equations?

The kinematic equations are a set of equations that describe the motion of an object with constant acceleration. Kinematics equations require knowledge of derivatives, rate of change, and integrals. To keep our focus on high school physics, we will not be covering integrals.

When we use the kinematic equations, we use specific notation to denote initial and final measurements. For example, when we have an initial velocity value, it is written as $ \Large\mathcal{v}_{\normalsize0} $. When we write the final velocity, we simply write $ \Large\mathcal{v} $, without a subscript. Therefore, the change in the velocity of the object is represented by the equation $ \Delta\Large\mathcal{v} = \mathcal{v} - \mathcal{v}_{\normalsize0} $. This notation also applies to displacement and time. This means $ {\large x}_0 $ is the initial position, $ {\large x} $ is the final position, $ {\large t}_0 $ is the initial time, and $ \large t $ is the final time. It’s important to remember that the initial time, $ {\large t}_0 $, will equal zero for kinematics equations. Because the time interval is $ \Delta\mathcal{t} $ and $ {\mathcal{t}}_0 = 0{\normalsize{s}} $, we know that $ \Delta\mathcal{t} = \mathcal{t} $.

How Can I Choose the Right Equation?

The kinematic equations can be applied to a variety of dimensional motion problems that consider the motion of an object with constant acceleration. When problem-solving, the formula we choose should include the unknown variable, as well as three known variables. Each of the equations is missing one variable. This allows you to identify what variable is not given or asked for in your problem, before selecting the equation that is also missing that variable. Let’s take a closer look.

Missing displacement? This formula is missing $ \Delta{\large x} $ and should be used for problems that do not include or ask for displacement.

$ \Large v = v_0 + \Large {at} $

Missing acceleration? This formula is ideal for problems that do not include or ask for acceleration. However, it’s important to remember that sometimes acceleration is indirectly referenced, such as when an object is in free fall (see below).

$ \Large \Delta \large x = \LARGE (\frac {v + v_0}{2}) \large t $

Missing final velocity? This formula can be used when the problem does not mention or ask for final velocity. Sometimes final velocity is indirectly referenced. When the problem includes, “comes to a stop” or “before stopping”, $ \large v = \normalsize 0 \textrm {m/s} $. This equation often utilizes the quadratic formula during problem-solving.

$ \Large{\Delta{\large x} = \LARGE v_{\normalsize{0}} \large t + \frac {1}{2}at^2} $

Missing time? When the problem does not give a time interval, choose this equation.

$ \LARGE v^2 = {v \normalsize{_0}}^2 + \Large 2a\Delta{\large x} $

Velocity, Acceleration, and Air Resistance

When we use the kinematics equations, we can make some mathematical assumptions. When an object in motion moves through the air, air resistance slows the object’s speed. Luckily, when we use the equations of motion, we assume air resistance is insignificant enough to ignore.

The second assumption we can make when using these equations involves acceleration. We already know that acceleration is constant for kinematics problems, which means that the average acceleration is equal to this value. Objects in free fall, or projectiles, all experience the same acceleration, regardless of their mass. This means that whenever an object is thrown, dropped, or falling, it moves with a constant downward acceleration of $ 9.81 \textrm {m/s}^2 $. It is important to remember that this value is a magnitude. If we assume upwards to be a positive direction or y value, then an object falling downward will have a negative acceleration of $ -9.81 \textrm {m/s}^2 $.

Because kinematics equations are used when the acceleration of the object is constant, we can use a simple equation to determine the average velocity of an object. To find the average velocity, simply add the initial velocity to the final velocity and divide by 2.

$ \LARGE v_{ \textrm {average}} = \bar{v} = \frac {v_1 + v_2}{2} $ This equation can only be used when acceleration is constant

When an object moves in a straight-line motion along the x-axis, we can use the displacement of an object and the time interval to determine the average velocity. This equation considers initial position, final position, and the time interval.

$ \LARGE v_{ \textrm {average}} = \bar{v} = \frac { x_2 - x_1}{ t_2 - t_1} = \frac {\Delta x}{\Delta t} $

You can use the slope of a line tangent to the position-time cure to determine the velocity of the object. The velocity of an object at a specific point in time is called instantaneous velocity. Graphing a position-time graph can help you to determine both instantaneous velocity and average velocity. Similarly, graphing a velocity-time graph can help you determine the instantaneous acceleration and average acceleration of an object because acceleration is the rate of change of velocity. However, when we apply this to the kinematics equations, which have constant acceleration, an object’s instantaneous acceleration will be equal to its average acceleration.

How to Derive the Kinematics Equations

The First Equation: $ \large v = v_0 + at $

1.) To begin deriving the first kinematic equation, we should first consider the definition of acceleration. $ \Large a = \frac {\Delta v}{\Delta t} $

2.) We know $ \Large \Delta v = v - v\normalsize{_0} $ , and when we plug that in, we get $ \Large a = \frac {v - v_0}{\Delta t} $

3.) If we solve for $ \Large v $, the equation becomes $ \Large v = v_0 + \Large a \Delta t $

$ \LARGE v = v_0 + at $

The Second Equation: $ \Delta x = (\frac {v + v_{\normalsize0}}{2}) t $

To derive this equation, we’ll consider a velocity-time graph with constant acceleration. The slope of a velocity graph can be interpreted as acceleration and the area under the graph is equal to the object’s displacement, $ \Delta x $ .

Here, the height of the blue rectangle is $ \large v_0 $ and the width is $ \large t $ so the area is equal to $ \large v_0 t $ .

The base of the red triangle is $ \large t $ and the height is $ \large v - v_0 $, so the area of the red triangle is $ \large \frac {1}{2} t (v - v_0) $ .

1.) When we sum the red and blue areas, we get $ \large \Delta x = v_0 t + \frac {1}{2} t (v - v_0) $

2.) When we distribute the factor of $ \large \frac {1}{2} t $ we get $ \large \Delta x = v_0 t + \frac {1}{2} vt - \frac {1}{2} v_0 t $

3.) We can simplify the equation by combining the initial velocity terms. $ \large \Delta x = v_0 t + \frac {1}{2} at^2 $

4.) Then, we further simplify the equation to get our second kinematic equation.

$ \LARGE \Delta x = (\frac {v + v_{\normalsize0}}{2}) t $

The Third Equation: $ \Delta x = v_0 t + \frac {1}{2} at^2 $

We can derive the third kinematic equation by plugging in the first kinematic formula into the second formula.

2.) Substitute the first kinematic formula for $ \large v $. $ \Large v = v\normalsize{_0} + \Large at $

3.) Once substituted, the equation becomes $ \large \frac {\Delta x}{t} = \frac {\Large (v_0 + at) + v_0}{2} $

4.) Expand the equation into $ \large \frac {\Delta x}{t} = \frac {\Large v_0}{2} + \frac {\Large at}{2} + \frac {\Large v_0}{2} $

5.) We can combine terms to simplify the equation into $ \large \frac {\Delta x}{t} = v_0 + \frac {\Large at}{2} $

6.) Finally, we can multiply both sides by time, $ \large t $, to generate our third kinematic equation.

$ \Large \Delta x = v_0 t + \frac {1}{2} at^2 $

The Fourth Equation: $ v^2 = {v_0}^2 + 2a \Delta x $

The fourth kinematic equation can be derived using the first and second kinematic equations.

1.) Start with the second kinematic formula. $ \large \Delta x = (\frac {v + v_{\normalsize0}}{2}) t $

2.) We can use the first kinematic formula, $ \Large \mathcal{v} = \mathcal{v}\normalsize{_0} + \large at $ , to solve for time. $ \large t = \frac {\Large v - v_{\normalsize0}}{a} $

3.) We can plug our expression for time into the second kinematic formula. $ \large \Delta x = (\frac {\Large v + v_{\normalsize0}}{2}) (\frac {\Large v - v_{\normalsize0}}{\Large a}) $

4.) Next, we can multiply the fractions to simplify. $ \large \Delta x = \Large (\frac {v^2 + {v_{\normalsize0}}^2}{\large 2a}) $

5.) When we solve for $ \large v^2 $, we get our fourth kinematic equation.

$ \LARGE v^2 = {v_0}^2 + 2a \Delta x $

Problem-Solving for Kinematic Equations

When solving kinematics problems, there are steps you can follow to help structure your thought process. After reading the problem, draw a diagram, and label the knowns and unknowns. Identify what you are being asked to find. Then, identify the variables the problem provides. Next, determine which equations connect your known variables to your unknown variable. Then, you can begin solving.

In this problem, motion only occurs in the downward direction, so we'll make downward the positive direction. $ g = 9.81 \textrm {m/s}^2 $

We’ll use the third Kinematics equation for this problem. $ \Delta x = v_0 t + \frac {1}{2} at^2 $

We know that $ \large v_0 = \normalsize 0 \textrm {m/s} $. We can eliminate terms involving $ \large x_0 = v_0 = \normalsize 0 $, let $ \large a = g $ and $ \large x = h $. After completing the substitutions, we get the equation $ \large h = \frac {1}{2} gt^2 $

Next, we need to create an expression for time.

$ \large v = gt $ can be rearranged into a function of time $ \large t = \frac {v}{g} $

Then we can plug our equation for time into our simplified equation for height to get our answer.

$ \Large h = \frac {1}{2} gt^2 = \frac {1}{2}g (\frac {v}{g})^2 = \frac {v^2}{2g} = \frac {(10 \textrm {m/s})}{2(9.81 \textrm {m/s}^2)} = \large 5.1 \textrm m $

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Wireless Photogate and Picket Fence

This problem only has motion occurring in the downward direction, so we can make downward the positive direction. $ g = 9.81 \textrm {m/s}^2 $

We know that $ \large x_0 = \normalsize 0 \textrm m $ and $ \large v_0 = \normalsize 0 \textrm {m/s} $.

We will use the kinematic equation $ \Delta x = v_0 t + \frac {1}{2} at^2 $.

We can eliminate terms, substitute $ \large a = g $, and assign $ \large h $ to height.

$ \large x = h $ $ \large h = (0)t + \frac {1}{2} gt^2 $

Plug in our values for gravity and time to create the final equation.

$ \Large h = \frac {1}{2} gt^2 = \frac {1}{2}(9.81 \textrm {m/s}^2)(1 \textrm s)^2 = 4.9 \textrm m $

Linear Motion in One Dimension

What are the given values? $ \large v_0 = 5 \textrm {m/s} $ $ \large v_0 = 0 \textrm {m/s} $ $ \large x - x_0 = 23 \textrm m $

We can rearrange the second kinematics formula $ \Delta x = (\frac {v + v_{\normalsize0}}{2}) t $

Substitute our equation for $ \Delta x $ and rearrange the equation. $ x - x_0 = \frac {1}{2}(v_0 + v) t $

Arrange the equation to solve for $ \large t $, and plug in our known values to solve.

$ \Large t = 2 (\frac {x - x_0}{v_0 + v}) = 2 (\frac {23 \textrm m}{5 \textrm {m/s} + 0 \textrm {m/s}}) = \large 11.6 \textrm s $

Dynamics Cart and Track System

Dynamics Cart and Track Systems are multi-purpose equipment sets that enable educators to demonstrate and teach a variety of complex physics topics in Kinematics and Dynamics. Since their invention, PASCO’s dynamics systems have become essential tools in physics labs around the world. Our dynamics systems are complete experiment solutions that enable students to observe, measure and analyze motion through hands-on experimentation. Whether you’re new to dynamics systems, or you’re looking to enhance your current systems’ capabilities, we offer a range of customizable dynamics systems to fit your budget and needs.

List the known values. $ \large v_0 = 0 \textrm {m/s} $ $ \large v = 13 \textrm {m/s} $ $ \large x - x_0 = 150 \textrm m $

Start with the first kinematics formula $ \large v^2 = {v_0}^2 \normalsize + 2a \Delta x $ and plug $ \large x - x_0 $ in for $ \Delta x $ .

$ \Large v^2 = {v_0}^2 + 2a (x - x_0) $

Rearrange the equation to solve for acceleration and plug in our known values, then solve.

$ \Large a = \frac {v^2 - {v_0}^2}{2(x - x_0)} = \frac {(13 \textrm {m/s})^2 - (0 \textrm {m/s})^2}{2(2.2 \textrm m)} = \large {38.0 \textrm {m/s}}^2 $

Linear Motion in the Second Dimension

Problems that include an inclined plane, ramp, or projectile will require us to split the motion into $ \small \overrightarrow{\normalsize x} $ and $ \small \overrightarrow{\normalsize y} $ vector components. Consider a jet with two perpendicular engines that propel it along the x and y axes. When only the vertically aligned engine is on, the jet is propelled along the y axis. Similarly, when only the horizontally aligned engine is on, the jet is propelled along the x axis. When both engines are on, the motion of the jet occurs in both the x and y directions, but neither x nor y affect one another. For this reason, we can analyze the $ \small \overrightarrow{\normalsize x} $ and $ \small \overrightarrow{\normalsize y} $ components for each vector separately, while recognizing that they are related by a shared $ \large t $ value. The directions of the components can be conveyed by assigning a plus or minus sign to each one.

We can represent $ \small \overrightarrow{\normalsize x} $ and $ \small \overrightarrow{\normalsize y} $ vector components with free body diagrams, a method commonly used to solve Newton’s laws problems. We will get more into this when we study dynamics. For now, it’s important to understand that some problems must be broken into x and y components. For example, when a cart goes down a ramp, it experiences motion in the x and y directions. Its motion depends on the net acceleration in the x-direction along the ramp. Because there is a component of the acceleration due to gravity that is accelerating the cart down the incline, part of the acceleration due to gravity is vertical. Below, you will find a vector diagram of a Smart Cart that breaks its motion into x and y components.

The magnitudes of the components of displacement along the axes are x and y. The magnitudes of the components of the velocity, $ \large v $ , are $ \large v_x = v\cos \Theta $ and $ \large v_y = v\sin \Theta $, where $ \large v $ is the magnitude of the velocity and theta is its direction.

Write down our known values $ \Theta = 45^{\circ} $ and $ \Delta x = 5.0 \textrm m $

We can start by taking the fourth Kinematics Equation $ {v_x}^2 = {v_{0x}}^2 + 2a_x( x - x_0) $ and rearranging it into $ {v_x}^2 - {v_{0x}}^2 = 2a_x \Delta x $

From the diagram, we see that the acceleration along the ramp is $ g \sin \Theta $. Plug $ g \sin \Theta $ and our known values into the equation and solve.

$ \Large {v_x}^2 = 2[(9.81 \textrm {m/s}^2)(\sin45^{\circ})] (5.0 \textrm m) = 8.3 \textrm {m/s} $

Wireless Smart Cart

The Wireless Smart Cart is a next-generation dynamics cart with built-in sensors that measure its position, velocity, acceleration, force and rotation. Fully functional with or without a track, the Wireless Smart Cart provides students with real-time motion data, interactive graphs, and intuitive analysis features on any device with a Bluetooth connection and PASCO software. From introductory experiments in linear and rotational motion, to Newton’s laws and multi-cart collisions, the Wireless Smart Cart enhances classic experiments and enables students to explore a new world of experimental opportunities.

Smart Cart Vector Display

Learn more about the right dynamics cart and track system setup for your classroom.

Projectile Motion

Projectile Motion occurs when the only acceleration experienced by an object in flight is caused by gravity, which pulls it downward. The object in motion is called a projectile, and its path is known as its trajectory. The horizontal distance covered by a projectile is called its range. The final location minus the initial location of a projectile is called its displacement, s. Two-dimensional projectile motion problems often assume air resistance and friction are negligible. Since the motions along a perpendicular axis are independent, we can analyze them separately by breaking them into their x and y components. When we apply the kinematic equations for each component, we use the x and y subscripts to denote each variable’s relation to the x- or y-axes. For both x and y components, acceleration is constant, which allows us to use the kinematic equations.

$ \Large a_y = g = -9.81 \textrm {m/s}^2 $

Here, we are assigning the upward y direction as positive, so a projectile experiencing the force of gravity, which pulls it in the downward y direction, will experience a negative velocity. Because a projectile is only affected by gravity, there is no acceleration in the x direction.

$ \large a_x = 0 \textrm {m/s}^2 $

A visual representation of the total displacement, $ \Large s $, of a projectile ball at a point along its path is shown above. The vector $ \Large s $, has components $ \Large x $ and $ \Large y $ along the axes. Its magnitude makes an angle $ \Theta $ with the horizontal. Notice that $ \Large s $ and $ \Large x $ can both denote displacement. Here, we are using $ \Large s $, to simplify each vector’s identification. We can use a series of steps to analyze projectile motion.

Step 1: Break the motion into horizontal and vertical components along the x- and y-axes.

$ \large A_x = A\cos\Theta $ $ \large A_y = A\sin\Theta $

The magnitudes of the components of displacement along the axes are $ \large x $ and $ \large y $. The magnitudes of the components of the velocity, $ \Large v $, are $ \Large v_x = v \normalsize \cos\Theta $ and $ \Large v_y = v \normalsize \sin\Theta $, where $ \Large v $ is the magnitude of the velocity and theta is its direction.

Step 2: Use the kinematic equations to analyze the components as two independent, one-dimensional motions. The kinematic equations for horizontal and vertical motion are below.

Horizontal Motion:

$ \large x = x_0 + v_{0x}t $

$ \Large v_x = v_{0x} $

Vertical Motion: $ a_y = g = -9.81 \textrm {m/s}^2 $

$ \large y = y_0 + \frac {1}{2}(v_{0y} + v_y)t $

$ \large v_y = v_{0y} - gt $

$ \large y = y_0 + v_{0y}t - \frac {1}{2} gt^2 $

$ \large {v_y}^2 = {v_{0y}}^2 - 2g(y - y_0) $

Step 3: Solve for the unknowns in the horizontal and vertical directions. Remember, both variables share the time variable.

Step 4: Recombine the two motions to determine the total displacements and velocity, $ \Large v $. Because these motions are perpendicular, we can determine these vectors by using the following vector summation methods.

$ \large A = \sqrt{{A_x}^2 + {A_y}^2} $ where $ \large \Theta = \tan^{-1} (A_y/A_x) $

We can use vector summation to determine the total displacement and velocity where $ \large \Theta $ is the direction of the displacement, and $ \large \Theta_v $ is the direction of the velocity.

$ \Large s = \sqrt{x^2 + y^2} $ and $ \large \Theta = \tan^{-1} (y/x) $

$ \Large v = \sqrt{{v_x}^2 + {v_y}^2} $ and $ \large \Theta_v = \tan^{-1} (v_y/v_x) $

When we consider the horizontal and vertical components of motion, we find the horizontal motion to be simple, because $ \large a_x = \normalsize 0 $ and $ \large v_x $ is constant. The velocity in the vertical direction begins to decrease as the object rises. At its height, the vertical velocity of the object is zero. As it falls back down, the vertical velocity increases in magnitude but in the opposite direction as its initial path. The x- and y-motions are recombined to find the total velocity at any given point during the projectile’s trajectory.

Projectile Launchers

We’re looking to find the distance in the horizontal direction and we’re given:

$ \large v_x = v_0 = \normalsize 10 \textrm {m/s} $ $ \large x = v_0t $ $ \large h = \normalsize 3.0 \textrm m $

We’ll use an equation in terms of y first. $ \large y = y_0 + v_{0y}t - \frac {1}{2}gt^2 $

Since the object starts from rest, we know that $ \large v_{0y} = 0 $ and $ \large v_y = -gt $

We can simplify the equation $ \large y = y_0 + v_{0y}t - \frac {1}{2}gt^2 $ because there is no initial velocity in the y direction.

When we plug in $ \Large v_{0y} = \normalsize 0 $ we get $ \large y = y_0 - \frac {1}{2}gt^2 $

We will use this equation in y to solve for the time. Substituting h for y gives us the equation $ \large h = h - \frac {1}{2}gt^2 $

When we rearrange to solve for time, the equation becomes $ \large t = \sqrt \frac {2h}{g} $

Then, we can calculate the displacement using our equation for $ \large t $, the $ \large x $ equation, $ \large x = v_0t $, and our known values.

$ \large x = v_0t = v_0 \sqrt \frac {2h}{g} = (10 \textrm {m/s}) \sqrt \frac {2(3.0 \textrm m)}{9.81 \textrm {m/s}^2} = 7.8 \textrm m $

We are looking to find $ \Large v_{0y} $ and our known values are $ \Large v_{0x} = \normalsize +3.0 \textrm {m/s} $ and $ \Large x = \normalsize 15 \textrm m $

We’ll use the horizontal component to find the shared value, time. Then we can rearrange the equation, $ \large x = x_0 + v_0t $ so that it becomes $ \large t = \frac {x}{v_{0x}} = \frac {15}{3.0 \textrm {m/s}} = \normalsize 5.0 \textrm s $

When we consider the vertical component of the motion, we can make the downward direction negative, which means $ \large a_y = \normalsize -9.81 \textrm {m/s}^2 $. We also know there isn’t any displacement in the vertical direction, so $ \large y = \normalsize 0 \textrm m $.

$ \Large y = v_{0y}t + \frac {1}{2} a_y t^2 $ becomes $ \Large 0 = v_{0y} t + \frac {1}{2}a_y t^2 $

Plug in our value for gravity and the calculated value for time to solve.

$ \Large v_{0y} = - \frac {1}{2} a_yt = - \frac {1}{2}(-9.81 \textrm {m/s}^2)(5.0 \textrm s) = 25.0 \textrm {m/s} $

Suggested Related Topics

Rotational motion.

The kinematics of rotational motion describes the relationships between angular velocity, rotation angle, angular acceleration, and time. Each of the kinematic variables for linear motion have a rotational motion counterpart. Like linear kinematic equations, the equations for rotational motion use subscripts to denote initial values, and exclude subscripts to denote final values. Below, you will find the equations for rotational motion and their translational, linear motion equations.

Straight-Line Motion of Charged Particles in Magnetic Fields

The first Newton’s law of motion states that if an object experiences no net force, then its velocity is constant. When a charged particle’s velocity is parallel to the magnetic field, it experiences no net force and moves in a straight line through space. This is known as straight-line motion. If the velocity vector is neither parallel nor perpendicular to the magnetic field, the velocity component that is parallel to the field will remain constant.

Rolling Objects with Different Moments of Inertia

One classic high school physics question involves two cylinders. The question states: You have two cylinders, one hollow and one solid, with identical masses and diameters. If you roll them both down a slope, which cylinder will reach the bottom first?

When we roll an object, its kinetic energy takes two forms: translational (motion in a straight line) and rotational (spinning). Translational energy with constant acceleration can be described using kinematic equations, but to answer our question, we must focus on rotational inertia. Rotational inertia depends not only on the mass and rotational speed of an object, but also on how mass is distributed around the axis of rotation. This means that a hoop will have more rotational inertia than a cylinder of equal mass at the same angular velocity, because the mass in the hoop is moving faster due to its greater distance from the axis of rotation.

The quantity of rotational inertia possessed by an object is called its moment of inertia. An object’s moment of inertia can also be considered to be a measurement of how “spread out” its mass it. The equation for an object’s moment of inertia is $ I = mr^2 $ , where $ I $ is equal to the moment of inertia, $ \large m $ is equal to the object’s mass, and $ \large r $ equals the radius. When two cylinders have different diameters but equal masses, the cylinder with the largest diameter will have a larger moment of inertia. Now, consider our original question. The hollow cylinder has the same mass and diameter as the solid cylinder, but its mass is more “spread out”, which causes it to have a bigger moment of inertia. So, does the magnitude of an object’s moment of inertia determine whether it will reach the bottom first? Not necessarily. In fact, if you roll the two cylinders down a ramp, the solid cylinder will reach the bottom first every time. It might seem counterintuitive, because the hollow cylinder has a larger moment of inertia, but a solid cylinder or sphere will always reach the bottom before a hollow object, regardless of diameter.

Planetary Orbits (Kepler’s Laws)

In the 16th-century, German astronomer, Johannes Kepler announced his three laws of planetary motion. The laws state: (1) All planets move about the Sun in elliptical orbits with the sun at one focus. (2) A radius vector joining any planet to the Sun sweeps out equal areas in equal lengths of time. (3) The squares of the sidereal periods of revolution of the planets are directly proportional to the cubes of their mean distances from the Sun.

These laws would eventually lend help to the development of Newton’s laws when he formulated the law of gravitation, in which he described the gravity between Earth and the Moon, as well as the Sun and the planets. Newton came to discover that the motion of bodies subject to central gravitational force don’t always follow the same elliptical orbits declared by Kepler’s first law. Instead, their paths are defined by the total energy of the body, allowing these bodies to follow hyperbolic or even parabolic orbits. Perhaps the most interesting aspect of Kepler’s laws is that they also apply to all other inverse-square-law forces, including electromagnetic forces within the atom (when proper allowances are made for relativistic and quantum effects).

Escape Velocity

Escape velocity is the speed that an object needs to be traveling to break free of a planet or moon’s gravity well and leave it without further propulsion. For example, an object on the moon must reach a speed of 2.38 km/sec to escape the moon’s gravity well. For comparison, an object on the Sun would need an escape velocity of 618 km/sec!

Satellite Orbits

Satellites are held within three types of Earth orbits in space. The high Earth orbit is farthest away from the Earth’s surface and is the location of many weather and communications satellites. The medium, or mid, Earth orbit includes navigation and most specialty satellites, and the low Earth orbit is where most scientific satellites orbit. The distance between the satellite and Earth’s surface, also known as the height of the orbit, determines the satellite’s orbital speed. The higher a satellite’s orbit is, the slower it will move. Other factors that influence a satellite’s orbit include eccentricity and inclination.

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Mathematics > Numerical Analysis

Title: an unsupervised deep learning approach for the wave equation inverse problem.

Abstract: Full-waveform inversion (FWI) is a powerful geophysical imaging technique that infers high-resolution subsurface physical parameters by solving a non-convex optimization problem. However, due to limitations in observation, e.g., limited shots or receivers, and random noise, conventional inversion methods are confronted with numerous challenges, such as the local-minimum problem. In recent years, a substantial body of work has demonstrated that the integration of deep neural networks and partial differential equations for solving full-waveform inversion problems has shown promising performance. In this work, drawing inspiration from the expressive capacity of neural networks, we provide an unsupervised learning approach aimed at accurately reconstructing subsurface physical velocity parameters. This method is founded on a re-parametrization technique for Bayesian inference, achieved through a deep neural network with random weights. Notably, our proposed approach does not hinge upon the requirement of the labeled training dataset, rendering it exceedingly versatile and adaptable to diverse subsurface models. Extensive experiments show that the proposed approach performs noticeably better than existing conventional inversion methods.

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Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are:

The symbols in the above equation have a specific meaning: the symbol d stands for the displacement ; the symbol t stands for the time ; the symbol a stands for the acceleration of the object; the symbol v i stands for the initial velocity value; and the symbol v f stands for the final velocity .

Applying Free Fall Concepts to Problem-Solving

There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows:

• An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object.
• If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s.
• If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity ( v f ) after traveling to the peak would be assigned a value of 0 m/s.
• If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height.

These four principles and the four kinematic equations can be combined to solve problems involving the motion of free falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized.  

Example Problem A

Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement ( d ) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles . For example, the v i value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above ). And the acceleration ( a ) of the shingles can be inferred to be -9.8 m/s 2 since the shingles are free-falling ( see note above ). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below.

The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d , v i , a , and t . Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables.

Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

-8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s 2 ) • (t) 2

-8.52 m = (0 m) *(t) + (-4.9 m/s 2 ) • (t) 2

-8.52 m = (-4.9 m/s 2 ) • (t) 2

(-8.52 m)/(-4.9 m/s 2 ) = t 2

1.739 s 2 = t 2

The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!  

Example Problem B

Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity ( v i ) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles . Note that the v f value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory ( see note above ). The acceleration ( a ) of the vase is -9.8 m/s 2 ( see note above ). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below.

The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are v i , v f , a , and d . An inspection of the four equations above reveals that the equation on the top right contains all four variables.

v f 2 = v i 2 + 2 • a • d

(0 m/s) 2 = (26.2 m/s) 2 + 2 •(-9.8m/s 2 ) •d

0 m 2 /s 2 = 686.44 m 2 /s 2 + (-19.6 m/s 2 ) •d

(-19.6 m/s 2 ) • d = 0 m 2 /s 2 -686.44 m 2 /s 2

(-19.6 m/s 2 ) • d = -686.44 m 2 /s 2

d = (-686.44 m 2 /s 2 )/ (-19.6 m/s 2 )

The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles . The next part of Lesson 6 provides a wealth of practice problems with answers and solutions.