# AP Physics 1: Kinematics Practice Problems with Answers

A complete set of multiple-choice questions about kinematics appearing in the AP Physics 1 exam is gathered here. For each section of kinematics, you can find here an answered multiple-choice question.

## AP Physics Kinematics Practice Problems

Problem (1): The position vs. time graph for a moving object along the $x$ axis is shown in the figure below. Which of the following graphs is the corresponding velocity vs. time graph? Solution: You can see similar questions to this one in all AP Physics kinematics practice problems, so we examine this more accurately.

The motion consists of three parts: two curves and one straight line.

The slope of a tangent line at each point of an $x-t$ curve gives the velocity of the object. Slopes with an acute angle ($\alpha<90^\circ$) indicate a positive velocity, and slopes with an obtuse angle ($\alpha>90^\circ$) show a negative velocity. Notice that these angles are measured relative to the positive $x$-axis.

In the first part (the blue curve), the slope at the starting point, which represents the initial velocity $v_0$, has an acute angle, indicating $v_0>0$. Throughout this time interval, all slopes have acute angles and are decreasing, suggesting that velocities are negative and decreasing in magnitude.

The middle part (the red straight line) has a constant slope in its time interval, indicating that the velocity must be constant during this time interval. The object continues to move at the same velocity as in the last second of the previous part.

The final part (the red curve) also has positive slopes but increases in angles, implying that all velocities are increasing in magnitude and positive as well.

Based on these explanations, it can be concluded that option (d) is correct. Hence, the correct answer is (d).

Problem (2): The velocity-versus-time graph of a moving object is shown in the figure below. Over time, which of the following is correct? (a) the velocity and acceleration are increasing.
(b) the velocity and acceleration are decreasing.
(c) the velocity increases but the acceleration decreases.
(d) the velocity decreases but the acceleration increases.

Solution: It is obvious that the object's velocity increases over time. The slope of a tangent line on a velocity vs. time graph always represents the magnitude and direction of acceleration. In this case, it can be observed that these slopes decrease from the initial time until reaching the final point on the graph. Therefore, it can be concluded that the object's acceleration is decreasing. Hence, the correct answer is (c).

Problem (3): Which of the following graphs shows the position vs. time graph for an object thrown vertically upward into the air? Solution: The basic position kinematic equation for an object moving at constant acceleration along a straight line is $x=\frac 12 at^2+v_0 t+x_0$ for an object that moves horizontally or $y=-\frac 12 gt^2+v_0t+y_0$ for vertical motion. In this problem, the object is thrown vertically upward, so its position at each moment is given by the second equation. Because of $-\frac 12 g$ in the second equation, the parabola has a concavity downward.

None of the given graphs above have the shape of a parabola, except graph (b).

Hence, the correct answer is (b).

Problem (4): The position vs. time graph of a moving object is shown in the figure below. How many times has the direction of the object changed?

(a) 1        (b) 2        (c)    3        (d) None. Solution: When the position vs. time graph is given and asked for the moments that the object changes its direction, use the following method: Find where the tangent line on the graph is horizontal. Recall that the slope of the tangent line on the x-t graph always shows the direction and magnitude of the velocity at any instant of time. So, where these tangent lines become zero (horizontal), the object reverses its direction. In the given graph, at times $t_1$ and $t_2$, the tangents are horizontal, so at least at these two points (positions), the direction of motion is reversed. But note that after time $t_2$, the position of the object does not change, which means the object is stopped. Thus, overall, the object has changed its direction only once. Hence, the correct answer is (a).

Problem (5): Consider the position vs. time graph for a moving object, as shown in the figure below. At which numbered points does the object have the greatest speed?

(a) 1        (b)    2        (c)    3        (d)    4 Solution: The slope of a tangent line at each point of an x-t graph gives the magnitude (speed) and direction of the velocity vector at that point. If we start drawing slopes at the origin, or point $4$, as we get closer to the top of the graph, the slopes get smaller and smaller. This continuous decrease in the slope of the curves indicates that the velocities are also decreasing until the top of the curve, where the velocity is totally zero. With this reasoning, we conclude that the slope at the numbered point $4$ must be the greatest.

Hence, the correct answer is (d).

Problem (7): The position vs. time graph for a moving object is shown in the figure below. How many times has the speed been zeroed and how many times has it reversed its direction, respectively?

(a) 1,1        (b) 2,2        (3) 1,2        (4) 2,1 Solution: The speed is zero where the tangent to the x-t curve becomes horizontal. As you can see, at three points, the tangents are horizontal. So, in these instants, the speed of the object is zero. Recall that the moment the speed of a moving object becomes zero, it changes direction or stops. Thus, at the three points shown above, the object changes its direction.

Hence, the correct answer is (b).

Problem (8): According to the figure below a car is moving at a constant rate along the $x$ axis. Which of the following velocity vs. time graphs describes the motion of the car? Solution: Take the positive to the right. The car moves to the left, so the direction of the velocity is negative. So far, the first choice is wrong since all the car's velocities have positive values, indicating moving to the right. The car starts to move non-stop to the left (in the negative direction). So, all the time, the car has negative velocity. Thus, choices (2) and (3) are also wrong.

Note that ''at a constant rate'' means constant acceleration.

On the other side, the acceleration is also toward the left. We know that the slope of the velocity-time graph gives the direction and magnitude of the acceleration. If the angle that the slope makes with the horizontal is obtuse ($\alpha >90^\circ$), then the direction of the acceleration is negative.

Hence, the correct answer is (d).

Problem (9): A car moving along the $x$ axis with a constant acceleration. If its velocity-versus-time graph is as below, Which of the following describes the motion of the car? Solution: First of all, choose a positive direction, say to the right. The graph shows us the initial velocity is positive. So, the car starts to move to the right initially. Thus, choices (3) and (4) are wrong. Since the graph does not intersect the time axis, means the velocity does not get zero, so the car continues to move to the right non-stop.

Now we must say something about the car's acceleration. Recall that the slope of a velocity-time graph gives the magnitude and direction of the acceleration. In this case, the slope makes an obtuse angle ($\alpha >90^\circ$) with the horizontal, which means the acceleration is negative or is toward the left. So, choice (2) is also incorrect.

Hence, the correct answer is (1).

Problem (10): The position of a moving object at any instant is given by equation $x=4t^2-6t+3$. What is the average velocity of the object between $t=1\,{\rm s}$ and $t=4\,{\rm s}$?
(a) 14        (b) 15        (c) 16        (d) 18

Solution: The average velocity in physics is defined as the ratio of displacement to the time interval taken. $v_{av}=\frac{x_2-x_1}{t_2-t_1}$ The positions at $4\,{\rm s}$ and $1\,{\rm s}$ are found as follows \begin{align*} x_1&=4(1)^2-6(1)+3 \\&=1\,{\rm m} \\\\\ x_4&=4(4)^2-6(4)+3 \\&=43\,{\rm m}\end{align*} Now use the definition of average velocity mentioned above to find it $v_{av}=\frac{43-1}{4-1}=14\,{\rm m/s}$
Hence, the correct answer is (a).

Method two: The position equation reveals that the motion has constant acceleration since it has the form of a quadratic equation. Comparing this with the standard position kinematic equation $x=\frac 12 at^2+v_0t+x_0$, we can find the acceleration $a$, the initial velocity $v_0$, and the initial position $x_0$ of the object as below \begin{gather*} \frac 12 a=4 \Rightarrow \quad a=8\,{\rm m/s^2} \\\\ v_0=-6\,{\rm m/s}\\\\ x_0=3\,{\rm m}\end{gather*} Substituting these kinematics variables into the velocity kinematics equation $v=v_0+at$, one can find the object's velocity at any instant of time $v=-6+8t$ By knowing the velocity versus time equation, we can determine the average velocity using the following formula $v_{av}=\frac{v_1+v_2}{2}$ The rest is left up to you as a practice problem.

Problem (11): A stone is thrown vertically upward into the air with an initial speed of $10\,{\rm m/s}$. After how many seconds does its velocity become $5\,{\rm m/s}$ downward?
(a) 0.5        (b) 1.5        (c) 2.5        (d) 2

Solution: Take positive as up so the initial velocity is positive $v_0=+10\,{\rm m/s}$. The velocity kinematic equation $v=v_0-gt$ gives us the velocity at any moment. We are asked for the time when the velocity is $5\,{\rm m/s}$ downward means $v=-5\,{\rm m/s}$ (Notice the negative sign). Substituting this information into the above velocity equation and solving for time gets $t$ \begin{gather*} v=v_0-gt \\\\ -5=10-(10)t \\\\ \Rightarrow \quad t=1.5\,{\rm s}\end{gather*} Hence, the correct answer is (b)

Problem (12): A car is moving at a constant speed of $54\,{\rm km/h}$ along a straight path. It applies brakes and stops after traveling a distance of $22.5\,{\rm m}$. How long is the stopping time?
(a) 6        (b) 5        (c) 2        (d) 3

Solution: In such AP kinematics questions, we must analyze only the braking stage. When the brakes are applied, the previous constant velocity becomes the initial velocity. After this moment, the velocity will change with time until it comes to a stop. So, during this stage, we have $v_0=54\,{\rm km/h}$, $v=0$, and $\Delta x=22.5\,{\rm m}$. Substituting these into the time-independent kinematic equation $v^2-v_0^2=2a\Delta x$ and solving for $a$ yields \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ 0-15^2=2a(22.5) \\\\ \Rightarrow \quad a=-5\,{\rm m/s^2}\end{gather*} where in above we converted the velocity units to the SI units ${\rm m/s}$ as below $54\,\frac{km}{h}=54\left(\frac{1000}{3600}\right)\frac ms=15\,\frac ms$ Now use the velocity kinematic equation $v=v_0+at$ and solve for $t$ \begin{gather*} v=v_0+at \\\\ 0=15+(-5)t \\\\ \Rightarrow \quad t=3\,{\rm s}\end{gather*} Hence, the correct answer is (d).

Quicker method: In all constant acceleration kinematics problems, we can use the following important formula $\Delta x=\frac{v_1+v_2}{2}\Delta t$ where $v_1$ and $v_2$ are the initial and final velocities during the accelerating stage, respectively. Here, $v_1=15\,{\rm m/s}$, $v_2=0$, and $\Delta x=22.5\,{\rm m}$. Substituting these into the above formula and solving for $t$ gives the same previous result.

Problem (13): The velocity of a car changes from $5\,{\rm m/s}$ to $7\,{\rm m/s}$ at a constant rate in a time interval of $5\,{\rm s}$. How long does the car move during this time period?
(a) 120 m        (b) 30 m
(c) 15 m         (d) 45 m

Solution: There are two methods to reach the answer. One longer way is, first, to find the car's acceleration, then use the equation $v=v_0+at$ and solve for $t$. Another much shorter way, which is suitable for AP Physics kinematics practice problems, is to use the formula below $\Delta x=\frac{v_1+v_2}{2}\Delta t$ Substituting the known numerical values into it and solving for $t$, we will find \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\Delta t\\\\ &=\frac{5+7}{2}(5) \\\\&=\boxed{30\quad {\rm m}}\end{align*} Hence, the correct answer is (b).

Problem (14): From the top of a height of 25 m, a bullet is thrown vertically into the air with an initial velocity of $20\,{\rm m/s}$. After how many seconds does the direction of motion of the bullet change?
(a) 2        (b) 3        (c) 4        (d) 5

Solution: Take the throwing point to be the origin, so $y_0=0$. The bullet rises high enough to reach zero speed. At this moment, it reverses its direction and starts to move down. During the ascent, the velocities at the initial and final points are known. So, solve the kinematic equation $v=v_0-gt$ for $t$. \begin{gather*} v=v_0-gt \\\\ 0=20-10t \\\\ \Rightarrow \quad \boxed{t=2\,{\rm s}}\end{gather*} Hence, the correct answer is (a).

Problem (15): In a projectile motion in a two-dimensional plane vertically, which of the following is correct:
(a) the vertical component of velocity is constant.
(b) the projectile's velocity is constant.
(c) the range of the projectile is independent of the throwing angle.
(d) the horizontal component of velocity remains constant.

Solution: The best approach to answering these kinds of kinematics questions in the AP physics exam is to write down the projectile motion formulas

(a) the vertical component of the velocity of a projectile is given by $v_y=v_0\sin\theta-gt$. As you can see, this component varies with time. So, this choice is wrong.

(b) The speed of the projectile definitely changes. Because initially it is thrown with some speed, and some later time at the highest point its velocity becomes zero. So, this choice is also wrong.

(c) The range of a projectile is found by $x=(v_0\cos\theta)t$. The range also changes with time.

(d) The horizontal component of the projectile's velocity is computed by $v_x=v_0\cos\theta$. As you can see, this component is independent of timeو so it remains constant during the motion.

Hence, the correct answer is (d).

Problem (16): A particle moves around a circle at a constant speed $v$. How many times is the change in velocity at points $A$ and $B$ relative to the magnitude of the velocity?
(a) zero    (b) 1        (c) $\frac 12$        (d) $\frac{\sqrt{3}}2$ Solution: The change in velocities at points $A$ and $B$ is also a vector quantity, $\Delta \vec{v}=\vec{v}_B-\vec{v}_A$. The problem has asked us to find the magnitude of this vector and compare it with the constant speed $v$ around the circle.

This change in velocity vector is a subtraction vector whose magnitude is given by the following formula $\Delta v=\sqrt{v_A^2+v_B^2-2v_A v_B \cos\theta}$ where $\theta$ is the angle between the two vectors. To find this angle, we must bring the velocity vectors to a point as shown below. Thus, the desired angle is $\theta=60^\circ$. Substituting $v_A=v_B=v$ into the above equation gives \begin{align*} \Delta v &=\sqrt{v_A^2+v_B^2-2v_A v_B \cos\theta} \\\\ &=\sqrt{v^2+v^2-2v^2\cos 60^\circ}\\\\ &=v\end{align*} The ratio wanted is found below $\frac{\Delta v}{v}=\frac{v}{v}=1$ Hence, the correct answer is (b).

More related articles:

Definition of a vector in physics
Practice problems on vectors and unit vectors.

Problem (17): A moving object revolves $90^\circ$ around a circle of radius $R$. How many times is the distance traveled by the object relative to the displacement?

(a) $\pi R/2$                (b) $R\sqrt{2}$
(c) $\pi/2$                    (d) $\pi \sqrt{2}/4$

Solution: This is refer to the vector problem in the AP physics exam. The distance traveled is simply one-fourth of the circumference of the circle. Thus, \begin{align*} D&=\frac 14 (circumference) \\\\ &=\frac 14 (2\pi R) \\\\ &=\frac{\pi R}{2}\end{align*} The object initially is at point $A$ and finally at point $B$.

From the origin, draw vectors that end at the locations of points $A$ and $B$. These vectors are called position vectors. The displacement vector is a vector that connects the initial point to the final point as shown (red dotted line) or $\vec{d}=\vec{R}_2-\vec{R}_1$ Thus, we must find the magnitude of this subtraction vector with the following standard formula \begin{align*} d&=\sqrt{R_1^2+R_2^2-2R_1 R_2 \cos\theta} \\\\ &=\sqrt{R^2+R^2-2R^2 \cos 90^\circ}\\\\ &=\sqrt{2R^2}\\\\&=R\sqrt{2}\end{align*} where in the above, $R$ and $\theta$ are the magnitudes of the position vectors and angle between them, respectively. Now construct their ratio as below: $\frac{D}{d}=\frac{\frac{\pi R}{2}}{R\sqrt{2}}=\frac{\pi \sqrt{2}}{4}$ Hence, the correct answer is (d).

Problem (18): An object is kicked horizontally with an initial velocity of $25\,{\rm m/s}$ and travels a horizontal distance of $50\,{\rm m}$ before landing. From what height was this ball thrown? (Ignore air resistance.)

(a) 20 m        (b) 30 m
(c) 25 m        (d) 45 m

Solution: This is another type of projectile motion problem that appears in all the AP Physics exams. First, list all the given data: horizontally kicked or thrown means the angle of the projectile is zero, $\theta=0$. The initial speed is $v_0=25\,{\rm m/s}$, and the horizontal distance traveled (which is the same range of projectile) is $50\,{\rm m}$.

To begin, calculate the total time the object is in the air using the projectile horizontal displacement formula $\Delta x=(v_0\cos\theta)t$. \begin{gather*} \Delta x=(v_0 \cos\theta)t \\\\ 50=(25\cos 0)t \\\\ \Rightarrow \quad t=2\,{\rm s}\end{gather*} where we used $\cos 0=1$. Now, use the projectile vertical displacement formula $y=-\frac 12 gt^2+(v_0\sin\theta)t+y_0$ and substitute the time $t$ found above into it and solve for $y$. But there is an important note about using this formula in all projectile motion problems and that is choosing a suitable coordinate system.

If we choose the initial throwing point as the origin, in this case, the initial position $y_0$ is set to zero, $y_0=0$ and this simplifies our calculations. Thus, \begin{align*} y&=-\frac 12 gt^2+(v_0\sin\theta)t+y_0 \\\\ &=-\frac 12 (10)(2)^2+0+0 \\\\ &=-20\,{\rm m}\end{align*} since $\sin 0=0$, we ignored that second term in the above and set it to zero. The negative indicates that the object will land $\boxed{20\,{\rm m}}$ below our hypothetical origin.

Hence, the correct answer is (a).

Problem (19): A $20\,{\rm g}$ ball is thrown with an initial velocity of $20\,{\rm m/s}$ vertically upward. How high does the ball reach after 2 seconds?

(a) 10 m        (b) 30 m
(c) 20 m        (d) 40 m

Solution: The object travels vertically, so it is better to use the freely falling kinematics equations. In this case, the vertical displacement is found using $\Delta y=-\frac 12 gt^2+v_0t$ If we choose the starting point as the origin, then $\Delta y$ gives us the vertical distance traveled by the object. Substituting the known numerical values into it, we will have \begin{align*} \Delta y&=-\frac 12 gt^2+v_0t\\\\ &=-\frac 12 (10) (2)^2+(20)(2) \\\\ &=\boxed{+20\,{\rm m}}\end{align*} Hence, the correct answer is (c)

Problem (20): From a height of $30\,{\rm m}$, a stone is thrown vertically downward with an initial speed of $5\,{\rm m/s}$. How many seconds after throwing the stone strike the ground?

(a) 1s            (b) 2s
(c) 3s            (d) 4s

Solution: Choose the starting point to be the origin, so the initial position becomes zero in the kinematics equations, $y_0=0$. The stone strikes 30 m below our origin, so its coordinate is $y=-30\,{\rm m}$. Note that the object moves down initially, so its velocity must be accompanied by a minus sign: $v_0=-5\,{\rm m/}$. The unknown is time $t$. Substitute this information into the vertical displacement kinematics equation below and solve for $t$ \begin{gather*} y=-\frac 12 gt^2+v_0t+y_0 \\\\ -30=-\frac 12 (10)t^2+(-5)t+0 \\\\ 5t^2+5t-30=0\end{gather*} This is a quadratic equation of the form $at^2+bt+c$ whose solutions is found as below $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ By putting the coefficients in the above equation, we have \begin{gather*} t=\frac{-5\pm\sqrt{5^2-4(5)(-30)}}{2(5)} \\\\ \boxed{t_1=2\,{\rm s}}\\\\ t_2=-3 \end{gather*} It is obvious that time cannot be negative. Hence, the correct answer is (b)

Problem (21): The position vs. time graph of a moving object is shown in the figure below. Is the object speeding up or slowing down at points $A$ and $B$, respectively? (a) Slowing down - speeding up.
(b) Both are slowing down.
(c) Both are speeding up.
(d) Speeding up - slowing down.

Solution: The terms ''speeding up'' or ''slowing down'' refer to an increase or decrease in the speed of a moving object, respectively. To determine this property at each point of an $x-t$ graph, we must determine the product $av$ at that particular point. If $av>0$ then the motion is speeding up, and if $av<0$ the motion is slowing down.

On the other hand, we know that the slope of a tangent line at each point on an $x-t$ curve gives the magnitude and direction of velocity at that point. If the slope makes an acute angle ($\alpha<90^\circ$) with the $+x$ direction, then $v<0$.

Opening up or down the curve (which is called concavity) also shows the sign (or direction) of the acceleration. For opening up concavity $a>0$ and for concavity down $a<0$. In this case, the points $A$ and $B$ lie on a curve that is opened downward so at both points we have $a>0$. The tangent line slope at point $A$ is an acute angle, so $v_A>0$ and it makes an obtuse angle at point $B$, so $v_B>0$.

Therefore, for point $A$ we have $a_A v_A>0$, so the object is speeding up. Similarly, at point $B$, $a_B v_B<0$, the object is slowing down.

Hence, the correct answer is (d)

Problem (22): A ball is thrown vertically upward into the air. What is the magnitude of the ball's acceleration just after leaving your hand?

(a) greater than g        (b) less than g
(c) equal g                        (d) zero.

Solution: Any motion in which the object is moving freely under the influence of gravity alone is called a freely falling motion. This definition is independent of the conditions of the initial motion. Objects thrown straight up or down and/or released from rest are all considered freely falling objects.

In this case, the only acceleration that the object experiences is the gravitational acceleration $g=9.8\,{\rm m/s^2}$. With these explanations, the correct answer is (c).

Author: Dr. Ali Nemati
Date Published: November 11, 2021