Kinematics Practice Problems
On this page, several problems related to Kinematics are given. The solutions to the problems are initially hidden, and can be shown in gray boxes or hidden again by clicking "Show/hide solution." It is advised that students attempt to solve each problem before viewing the answer, then use the solution to determine if their answer is correct and, if not, why. Remember to include units on all final answers.
Content that appears in a box similar to this is content that applies only to the AP C curriculum. AP B and AICE students can skip any content contained in these boxes.
- A rock is thrown straight upward off the edge of a balcony that is 5 m above the ground. The rock rises 10 m, then falls all the way down to the ground below the balcony. What is the rock's displacement?
Show/hide solutionDisplacement deals only with the initial and final positions of the rock. The rock's initial position is the balcony, and its final position is the ground 5 m below. So, the rock's displacement is 5 m downward.
- A child walks 5 m east, then 3 m north, then 1 m east.
a.) What is the magnitude of the child's displacement?
b.) What is the direction of the child's displacement?
Show/hide solutiona.) The child's total distance is 5 + 3 + 1 = 9, however the displacement is the straight-line distance from the child's initial position to its final position. It is best to start by drawing a diagram.
As you can see in this diagram, the displacement is equivalent to the hypotenuse of a right triangle whose legs are 6 m and 3 m long. So, we can calculate the magnitude of the displacement using the Pythagorean theorem:
d = sqrt(Δx2 + Δy2) = sqrt(62 + 32) = sqrt(45) = 6.7 m
b.) The child's direction is equal to the angle formed between d and Δx, which we will call θ. Here we must use some trigonometry. So, we will use the fact that tanθ = opposite/adjacent to find θ:
tanθ = Δy/Δx
θ = tan-1(Δy/Δx) = tan-1(3/6) = tan-1(0.5) = 26.5651o ~ 26.5o north of east
Notice the direction takes the form of an angle, 26.5o, and a reference for the angle, north of east. The second part is given so that it is clear what axis the angle is being measured from and in what direction from the axis. North of east means that the angle is measured from the east, or positive x, axis, and is north, the positive y direction, from this axis.
- An athlete runs exactly once around a circular track with a total length of 500 m. Find the runner's displacement for the race.
Show/hide solutionSince the athlete runs exactly once around the circular track, that means that he/she finishes at the same point where he/she started, meaning the initial and final positions are the same. Thus, the displacement is zero, and the total length of the track is irrelevant to the question.
Speed and Velocity
- If the child from problem 2 completes his journey in 20 seconds, what is the magnitude of his average velocity?
Show/hide solutionWe found in problem 2 that the child's displacement is 6.7 m, so the magnitude of his average velocity is:
vavg = Δx/Δt = 6.7/20 = 0.34 m/s
- If the runner from problem 3 runs the lap in 1 minute 18 seconds, find his/her average speed and the magnitude of his/her average velocity.
Show/hide solution1 minute 18 seconds is the same as 78 seconds. Average speed is the total distance, not displacement, divided by the time, and the track is 500 m long, so the average speed is:
500/78 = 6.4 m/s
Average velocity is displacement divided by time. We found in problem 3 that the runner's displacement was 0 m, so:
vavg = 0/78 = 0 m/s
- a.) Is it possible to move with constant speed but not constant velocity?
b.) Is it possible to move with constant velocity but not constant speed?
Show/hide solutiona.) Yes. For example, if a car travels at constant speed while going around a curve in the road, its speed remains constant. But since velocity also includes direction, and the car's direction is changing, the velocity is not constant.
b.) No. Since speed is the magnitude of velocity, any object with constant velocity must have constant speed.
- A car drives in a straight line at a constant speed of 60 miles per hour for 5 seconds. Find its acceleration.
Show/hide solutionSince the car's velocity is constant, there is no acceleration. So, a = 0.
- A remote control car is driven along a straight track at 2 m/s. The child controlling the car then activates the toy's turbo mode so that, 3 seconds later, the car's speed is 3.2 m/s. Find its average acceleration.
Show/hide solutionThe car has an initial velocity vi = 2 m/s and a final velocity vf = 3.2 m/s, over a time interval Δt = 3 s. So:
aavg = (vf - vi)/Δt = (3.2 - 2)/3 = 1.2/3 = 0.4 m/s2
- Shortly after, the remote control car in the previous example exits turbo mode, slowing from 3.2 m/s back to 2 m/s in 2 seconds. Find the car's average acceleration over this interval.
Show/hide solutionThe car now has an initial velocity vi = 3.2 m/s and a final velocity vf = 2 m/s, over a time interval Δt = 2 s. So:
aavg = (vf - vi)/Δt = (2 - 3.2)/2 = -1.2/2 = -0.6 m/s2
Note that the acceleration is negative because the car slowed down while moving in the positive direction, meaning the acceleration effectively occured in the direction opposite to the velocity.
Uniform Acceleration w/ the Big Five
- A particle moves along the x-axis with an initial velocity of 4 m/s and constant acceleration. After 3 seconds, its velocity is 14 m/s. How far did it travel during this interval?
Show/hide solutionFirst, let's list the givens and unknowns:
Given: vi = 4 m/s, vf = 14 m/s, Δt = 3 s
Unknown: Δx = ?
Since we know vi, vf, and Δt, and need to know Δx, but a is not specifically given, we must use the Big 5 equation that uses all values except a. This is Big 5 number 1:
Δx = 1/2(vi + v)t = 0.5(4 + 14)3 = 0.5 * 18 * 3 = 9 * 3 = 27 m
- A car is initially moving at 10 m/s and accelerates at a constant rate of 2 m/s2 for 4 seconds, in a straight line. How far did the car travel during this time?
Show/hide solutionGiven: vi = 10 m/s, a = 2 m/s2, Δt = 4 s
Unknown: Δx = ?
The value that is not given or asked for is v, so we must use the equation that does not contain v. This turns out to be Big 5 number 3:
x = 1/2at2 + vit + xi = 0.5 * 2 * 42 + 10 * 4 = 16 + 40 = 56 m
Note that since we are not given an initial position and are asked for the displacement, not the final position, we simply assume that the initial position is 0, which makes the displacement equal to the final position.
- A rock is dropped from a cliff that is 80 m above the ground. If the rock hits the ground with a velocity of 40 m/s, what acceleration did it undergo?
Show/hide solutionSince the rock is dropped, we know that it began at rest, so the initial velocity is 0.
Given: vi = 0, Δx = 80 m, v = 40 m/s
Unknown: a = ?
The value that is neither given nor asked for is Δt, so we must use the equation which does not contain Δt. This is Big 5 number 5:
v2 = vi2 + 2a(x - xi) = 2a(x - xi) = 2aΔx
a = (v2)/(2Δx) = (402)/(2 * 80) = 1600/160 = 10 m/s2, toward the ground.
- A rock is dropped 80 meters from a cliff. How long does it take to reach the ground?
Show/hide solutionUnless a problem specifically states otherwise, we can assume the object is being dropped on earth, so we know g = 9.81.
Given: vi = 0 m/s, Δy = 80 m, a = 9.81 m/s2
Unknown: Δt = ?
Since we are not given nor asked for v, we can use the Big 5 number 3, adapted to the y direction:
y = 1/2at2 + vit + yi
Notice that here, we can either choose to set yi = 0 and y = -80, or y = 0 and yi = 80. Both will yield the same answer, but here we will choose the first option so that two terms are equal to zero on the right side.
y = 1/2at2
t2 = 2y/a
t = sqrt(2y/a) = sqrt(2 * -80/-9.81) = 4.04 s
If we needed to do this math without a calculator, we would substitute -10 instead of -9.81 for a, yielding an answer of 4 s. Both answers would be accepted on either section of either AP Physics exam.
- A ball is thrown straight up with an initial speed of 20 m/s. How high will the ball travel?
Show/hide solutionAt first it may look like there are two variables that are neither given nor asked for, t and v. However, we know that at the highest point in a free-falling object's arc, the vertical velocity becomes 0 for an instant. Therefore, v = 0, and since t is the missing variable, we use Big 5 number 5:
Given: vi = 20 m/s, a = -9.81 m/s2, v = 0 m/s
Unknown: Δy = ?
v2 = vi2 + 2aΔy = 0
-2aΔy = vi2
Δy = vi2/(-2a) = 202/(-2 * -9.81) = 20.4 m
Or, if -10 is used instead of -9.81:
202/(-2 * -10) = 202/20 = 20 m
- One second after being thrown straight down, a rock is falling at 20 m/s. How fast will it be falling 2 seconds later?
Show/hide solutionSince the initial velocity is downward, we give it a negative sign.
Given: vi = -20 m/s, a = -9.81 m/s2, Δt = 2 s
Unknown: v = ?
Since x is irrelevant to the problem, we use Big 5 number 2:
v = at + vi = -9.81 * 2 - 20 = -19.62 - 20 = -39.62 m/s
-10 * 2 - 20 = -20 - 20 = -40 m/s
Note that this number is negative to indicate that the rock's velocity is in the downward, or negative, direction.
- An object is thrown straight upward with an initial speed of 8 m/s and strikes the ground 3 seconds later. What height was the object thrown from?
Show/hide solutionGiven: a = -9.81 m/s2, vi = 8 m/s, Δt = 3 s
Unknown: yi = ?
The variable that is niether given nor asked for is v, so we will use Big 5 number 3, setting y equal to 0, since the final height is the ground:
y = 1/2at2 + vit + yi
yi = y - 1/2at2 - vit = -0.5 * -9.81 * 32 - 8 * 3 = 20.145 m
-0.5 * -10 * 32 - 8 * 3 = 5 * 9 - 24 = 45 - 24 = 21 m
- A tennis ball is thrown horizontally with an initial speed of 10 m/s. If it hits the ground after 4 seconds, how far did it drop before hitting the ground?
Show/hide solutionGiven: vx,i = 10 m/s, vy,i = 0 m/s, Δt = 4 s, ay = 9.81 m/s2
Unknown: Δy = ?
We are looking for information in the vertical direction, so that will be our primary focus. If we call yi, the initial height, zero, then y will be our answer. It turns out that the initial horizontal velocity is irrelevant and we can use the vertical information in Big 5 number 3.
y = 1/2at2 + vy,it + yi = 0.5 * -9.81 * 42 + 0 * 4 + 0 = 0.5 * -9.81 * 16 = -78.48 m
0.5 * -10 * 42 = -5 * 16 = -80 m
- A ball is thrown horizontally from a height of 100 m with an initial speed of 15 m/s. How far does it travel horizontally in the first 2 seconds?
Show/hide solutionGiven: yi = 100 m, vx,i = 15 m/s, vy,i = 0 m/s, ay = -9.81 m/s2, Δt = 2 s
Unknown: Δx = ?
Here, we are looking for information in the horizontal direction. In the notes, we derived the equation Δx = vx,it. Since we know vx,i and t, we can simply plug the numbers into this formula and have our answer:
Δx = vx,it = 15 * 2 = 30 m
- A javelin travels in a parabolic arc for 6 seconds before hitting the ground. Compare its horizontal velocity 1 second after being thrown to its horizontal velocity 4 seconds after being thrown.
Show/hide solutionAssuming air resistance is negligible, there is no acceleration in the horizontal direction during projectile motion. Therefore, the javelin's horizontal velocity cannot change at any time during the flight, so its horizontal velocities 1 second and 4 seconds after being thrown are the same.
- A circus performer is launched out of a cannon with a launch angle of 30o and an initial velocity of 40 m/s. How long after launch will it take for the performer to reach the top of his trajectory, and how high is this point?
Show/hide solutionGiven: θ = 30o, vi = 40 m/s, ay = -9.81 m/s2
Unknown: tymax = ?, ymax = ?
Since the performer is launched at an angle, the first thing we need to do is resolve his initial velocity into its horizontal and vertical components. Since this problem deals only with the vertical direction, we need only find the vertical component:
sinθ = vy,i/vi
vy,i = visinθ = 40sin(30o) = 20 m/s
We know that at the highest point in any projectile's trajectory, its vertical velocity becomes 0 during that instant. So if we set the equation for a projectile's vertical velocity equal to 0, we can solve for the time at that point:
vy = ayt + vy,i = 0
ayt = -vy,i
t = -vy,i/ay = -20/-9.81 = 2.04 s
-20/-10 = 2 s
Now that we know the time after launch at the performer's maximum height, we can calculate his maximum height using the equation for vertical position, assuming his initial height is the ground, or 0 m:
y = 1/2at2 + vy,it + yi = 0.5 * -9.81 * 2.042 + 20 * 2.04 = 20.39 m
0.5 * -10 * 22 + 20 * 2 = -5 * 4 + 40 = -20 + 40 = 20 m
- A cannonball is fired from the ground at a 30o angle with an initial speed of 60 m/s. If the cannonball lands back at the same height it was launched from:
a.) For how long will the cannonball be in the air?
b.) How far will it travel horizontally?
Show/hide solutionGiven: θ = 30o, vi = 60 m/s, ay = -9.81 m/s2
a.) Unknown: Δt = ?
In order to find how long the cannonball travels, we could determine the time at which its height is 0, however this requires us to solve a quadratic equation, which is too difficult without a calculator. The alternate solution, since the cannonball landed at the same height it was launched from, is to find when it reaches its maximum height, as in the previous problem, then double that number. The reason this works is because the vertical acceleration is constant, so it will take the cannonball the same amount of time to go up a given distance as it will to go down the same distance. Before we can find this, however, we must again find the vertical component of the initial velocity:
sinθ = vy,i/vi
vy,i = visinθ = 60sin(30o) = 30 m/s
vy = ayt + vy,i = 0
ayt = -vy,i
t = -vy,i/a = -30/-9.81 = 3.06 s
3.06 * 2 = 6.12 s
-30/-10 = 3 s
3 * 2 = 6 s
b.) Unknown: Δx = ?
Now that we know how long the cannonball is in the air, it's a simple matter of using our equation for horizontal displacement. The only difficulty is that we must first find the horizontal component of the initial velocity:
cosθ = vx,i/vi
vx,i = vicosθ
Δx = (vicosθ)t = 60cos(30o)6.12 = 318 m
60cos(30o)6 = 312 m
Note that you would generally not be asked to find cos(30o) on the multiple choice section of the AP exam. This is because cos(30o) = sqrt(3)/2, which is not easily calculated by hand.
Kinematics w/ Graphs
- An object's position during a 10 second time interval is shown by the graph below:
a.) Determine the object's total distance traveled and displacement.
b.) What is the object's velocity at the following times: t = 1, t = 3, and t = 6.
c.) Determine the object's average velocity and average speed from t = 0 to t = 10.
d.) What is the object's acceleration at t = 5?
Show/hide solutiona.) The total distance traveled by the object is the sum of all the distances it traveled during the time interval. In the first two seconds it traveled 3 m. Then it traveled 0 m in the next two seconds. Then over the next five seconds, the object moved 5 m, then remained at rest. so the total distance is 3 + 5 = 8 m.
The displacement of the object is simply the final position minus the initial position, or -2 - 0 = -2 m.
b.) Notice that each of these points is in the middle of a line segment on the graph. Because of this, the instantaneous velocity at these points is the same as the average velocity over the time intervals represented by each segment, so:
v(t) = (xf - xi)/(tf - ti)
v(1) = (3 - 0)/(2 - 0) = 3/2 = 1.5 m/s
v(3) = (3 - 3)/(4 - 2) = 0/2 = 0 m/s
v(6) = (-2 - 3)/(9 - 4) = -5/5 = -1 m/s
Notice that the formula (xf - xi)/(tf - ti) is the same as the slope formula for this graph. The velocity at any point on a position vs. time graph is simply the slope of the graph at that point.
By this definition, we also know that the velocity of any position function is its derivative with respect to time. You can also go from a velocity function to a position function using integration.
c.) Average velocity is displacement divided by time. We found in part a that the object's displacement is -2 m, so:
vavg = -2/10 = -0.2 m/s
Average speed is total distance divided be time, and we found in part a that the object's total distance traveled is 8 m. so:
8/10 = 0.8 m/s
d.) We determined in part b that the object's velocity is represented by the slope of the line segment on the graph. Since the slope of this segment is constant, the object's velocity at t = 5 is constant. Since constant velocity means there is no acceleration, a = 0.
- An object's velocity during a 10 second time interval is shown by the graph below:
a.) Determine the object's total distance traveled and displacement.
b.) At t = 0, the object's position is x = 2 m. Find the object's position at t = 2, t = 4, t = 7, and t = 10.
c.) What is the object's acceleration at the following times: t = 1, t = 3, and t = 6.
d.) Sketch the corresponding acceleration vs. time graph from t = 0 to t = 10.
Show/hide solutiona.) Recall that the equation for velocity is v = x/t. If we solve this for x, we get x = vt. Notice that this is the same as the area of a rectangles whose sides are lengths v and t, so we can determine that the displacement is the area enclosed by the velocity vs. time graph. So, we will find the area of each section under the graph:
The total distance traveled by the object is simply the sum of all these areas: 3 + 6 + 4.5 + 2 + 2 = 17.5 m
The displacement is found in a similar fashion, except areas below the x-axis are considered negative: 3 + 6 + 4.5 - 2 - 2 = 9.5 m
Interestingly enough, the area enclosed by any function can be represented by a definite integral. For example, if this graph were defined as a function v(t), then the displacement would be the integral from 0 to 10 of v(t)dt, and the total distance traveled would be the integral from 0 to 10 of |v(t)|dt
b.) The position of the object at a given point in time can be found in much the same way we found the displacement in part a, except this time we must also add in the initial value given. So:
x(2) = 2 + 3 = 5 m
x(4) = 2 + 3 + 6 = 11 m
x(7) = 2 + 3 + 6 + 4.5 = 15.5 m
x(10) = 2 + 3 + 6 + 4.5 - 2 - 2 = 11.5 m
Notice that this can also be done by adding the integral from 0 to t of v(t)dt to the initial value of 2.
c.) Like velocity in part b of problem 22, the instantaneous acceleration at any point along one of the graph's line segments is the same as the average acceleration across that line segment. The formula for acceleration is aavg = Δv/Δt = (vf - vi)/(tf - ti), so:
a(t) = (vf - vi)/(tf - ti)
a(1) = (3 - 0)/(2 - 0) = 3/2 = 1.5 m/s2
a(3) = (3 - 3)/(4 - 2) = 0/2 = 0 m/s2
a(6) = (-2 - 3)/(9 - 4) = -5/5 = -1 m/s2
Similarly to the relationship between velocity and position, the formula for acceleration is the same as the slope formula for a velocity vs. time graph. So, we can say that the slope of any velocity vs. time graph is its acceleration.
Notice that this definition defines acceleration as the derivative of velocity. So, it is true that for any velocity function v(t), its derivative is an acceleration function a(t). Also, integration can be used to go from an acceleration function to a position function.
d.) We know that the acceleration along each line segment of this velocity vs. time graph is equal to the slope of the line segment. We determined these slopes in part c, so the acceleration graph would look like:
This graph uses horizontal lines instead of points to represent that the acceleration is defined at that value at any point along that section. The open circles at the end of each line segment simply indicate that at those time values, acceleration is not defined at either value represented by the horizontal lines. At these points, acceleration is undefined because it changes instantaneously from one value to the next, which cannot be represented numerically.
- An object's position during a given time interval is shown by the graph below:
a.) At which of the marked points is the object's velocity the greatest? The least?
b.) Is the object's acceleration positive or negative between points A and B?
c.) Suppose this curve can be modeled by the function x(t) = t3 - 9.5t2 + 23t - 9. Find the object's velocity and acceleration at t = 1, t = 3, and t = 5.
d.) Using the function from part c, determine the object's maximum and minimum positions and velocities within the interval from t = 1 to t = 6.
Show/hide solutiona.) Remember from problem 22 that velocity is the slope of a position vs. time graph such as this. By looking at lines tangent to the curve, we can see which point has the highest and lowest slope:
Looking at the red tangent lines, we can immediately eliminate point B as a candidate for both the maximum and minimum velocity, as its tangent is horizontal and thus has a slope of 0. Point C is the only marked point whose tangent line has a negative slope, so point C has the lowest velocity. Looking at points A and D, point A's tangent line has a steeper positive slope so point A has the highest velocity.
b.) We know that acceleration is a change in velocity, so by asking whether acceleration is positive or negative, we are asking if the velocity is increasing or decreasing. Since velocity is the slope of this graph, we must determine how the slope of the curve is changing between points A and B. Looking at the diagram in part a, we see that the slope at point A is positive, and the slope at point B is 0. As such, the slope, and thus the velocity, must be decreasing. Therefore, the object's acceleration is negative in this interval.
c.) We know from problem 22 that velocity is the derivative of position, and from problem 23 that acceleration is the derivative of velocity. So, we will begin by differentiating the position function twice:
x(t) = t3 - 9.5t2 + 23t - 9
v(t) = 3t2 - 19t + 23
a(t) = 6t - 19
Now that we know the velocity and acceleration functions, all that is left is to plug the values of t into these functions and simplify:
v(1) = 3 * 12 - 19 * 1 + 23 = 3 - 19 + 23 = 7 m/s
v(3) = 3 * 32 - 19 * 3 + 23 = 27 - 57 + 23 = -7 m/s
v(5) = 3 * 52 - 19 * 5 + 23 = 75 - 95 + 23 = 3 m/s
a(1) = 6 * 1 - 19 = 6 - 19 = -13 m/s2
a(3) = 6 * 3 - 19 = 18 - 19 = -1 m/s2
a(5) = 6 * 5 - 19 = 30 - 19 = 11 m/s2
d.) Thinking logically about the graph, the possible candidates for maximum and minimum position are at the end points of the interval and at the spots, like point B, where the slope of the graph is 0. So, first we set the velocity function from part c equal to 0 and solve for t:
v(t) = 3t2 - 19t + 23 = 0
t = 1.63008 s or t = 4.70326 s
Note that this was solved using a graphing calculator. The AP exam will not ask you to solve a quadratic this complicated by hand, however you may have to solve a simpler function using the quadratic formula. Also, we keep as many decimal places as we can at this stage in order to maintain accuracy. Now that we know all the possible times at which the position could be at a maximum or minimum within the interval, we simply plug these t values into x(t). Don't forget to check the end points:
x(t) = t3 - 9.5t2 + 23t - 9
x(1) = 5.5 m
x(1.63008) = 7.58 m
x(4.70326) = -6.93 m
x(6) = 3 m
We see that the minimum position is -6.93 m, and the maximum position is 7.58 m. Finding the maximum and minimum velocities is achieved in the same manner, except we set the acceleration function equal to 0 and plug the t values into the velocity funtion:
a(t) = 6t - 19 = 0
6t = 19
t = 19/6 = 3.16667 s
v(t) = 3t2 - 19t + 23
v(1) = 7 m/s
v(3.16667) = -7.08 m/s
v(6) = 17 m/s
So the minimum velocity is -7.08 m/s, and the maximum velocity is 17 m/s.