IB Physics Quiz: Understand Kinematics

20 questions · exam conditions

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Understand KinematicsQuestion 1 of 20

A stone is thrown vertically upwards from the ground. It reaches a maximum height $H$ and falls back to the ground. Air resistance is negligible. Taking the upward direction as positive, which statement is correct regarding the signs of displacement, velocity, and acceleration while the stone is in the air?

The acceleration is always negative, while the velocity is first positive then negative.

The displacement is always positive, while the velocity is always positive.

The product of velocity and acceleration is always negative.

The acceleration is first positive then negative, while displacement is always positive.

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IB Physics Quiz

IB Physics Quiz: Understand Kinematics

Practice Understand Kinematics in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Understand Kinematics, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The acceleration is always negative, while the velocity is first positive then negative. (correct answer)
The displacement is always positive, while the velocity is always positive.
The product of velocity and acceleration is always negative.
The acceleration is first positive then negative, while displacement is always positive.

Explanation: Acceleration due to gravity is always directed downwards, so it is always negative in this coordinate system. The stone's velocity is initially positive (upwards), becomes zero at the maximum height, and is negative (downwards) on its return trip. Therefore, statement A is correct. Distractor B is wrong because velocity becomes negative. Distractor C is wrong because on the way down, both velocity and acceleration are negative, making their product positive. Distractor D is wrong because acceleration is constant and always negative.

Question 2

An object moves along a semi-circular path of radius $r$ from point P to point Q in time $T$ . What is the magnitude of the average velocity divided by the average speed?

$\frac{\pi}{2}$
$\frac{2}{\pi}$ (correct answer)
1
$\pi$

Explanation: The distance travelled along the semi-circular path is the arc length,

d = \pi r

. The average speed is

\frac{\text{distance}}{T} = \frac{\pi r}{T}

. The displacement is the straight-line distance from P to Q, which is the diameter of the circle,

s = 2r

. The magnitude of the average velocity is

\frac{|\text{displacement}|}{T} = \frac{2r}{T}

. The required ratio is

\frac{\text{magnitude of average velocity}}{\text{average speed}} = \frac{2r/T}{\pi r/T} = \frac{2}{\pi}

. Distractor A is the reciprocal of the correct answer. Distractor C confuses distance with displacement. Distractor D makes an error in calculating displacement or distance.

Question 3

A boat, which has a speed of $v$ in still water, is to cross a river of width $W$ . The river flows at a constant speed $u$ . To cross the river in the shortest possible time, the boat should be steered in which direction relative to the river banks?

Directly upstream to counteract the flow partially.
At an angle upstream such that its resultant velocity is perpendicular to the banks.
Directly perpendicular to the river banks. (correct answer)
Directly downstream to take advantage of the flow.

Explanation: The time to cross the river is determined by the component of the boat's velocity that is perpendicular to the river flow. Time

t = \frac{W}{v_{\perp}}

. To minimize the time

t

, the perpendicular component of the velocity,

v_{\perp}

, must be maximized. The boat's velocity relative to the water is

v

. This component is maximized when the boat is pointed directly across the river, perpendicular to the banks, making

v_{\perp} = v

. Distractor B describes the path to minimize the distance travelled downstream (to go straight across), not the time taken.

Question 4

A ball is thrown vertically upwards and returns to its starting point. Air resistance is significant. How does the time taken to rise to its maximum height, $t_{up}$ , compare with the time taken to fall back, $t_{down}$ ?

$t_{up} < t_{down}$ (correct answer)
$t_{up} = t_{down}$
$t_{up} > t_{down}$
The relationship depends on the initial speed of the ball.

Explanation: On the way up, both gravity and air resistance act downwards. The net downward acceleration

a_{up}

is greater than

g

. On the way down, gravity acts downwards but air resistance acts upwards, so the net downward acceleration

a_{down}

is less than

g

. Since the ball travels the same vertical distance up and down, but the average magnitude of acceleration is greater on the way up, the time taken for the upward journey must be shorter.

t_{up} < t_{down}

. Distractor B is only true in a vacuum. Distractor C has the reasoning reversed. Distractor D is incorrect because while the values of the times depend on the speed, the inequality

t_{up} < t_{down}

holds regardless of the initial speed.

Question 5

A projectile is launched with a certain speed $u$ . For a given horizontal range $R$ , which is less than the maximum possible range, it is known that two launch angles, $\theta_1$ and $\theta_2$ , will result in hitting the target. What is the relationship between these two angles?

$\theta_1 + \theta_2 = 45°$
$\theta_1 + \theta_2 = 90°$ (correct answer)
$\theta_1 = \theta_2$
$\theta_2 = 2\theta_1$

Explanation: The range equation for a projectile is

R = \frac{u^2 \sin(2\theta)}{g}

. For a fixed

u

and

R

, we need

\sin(2\theta_1) = \sin(2\theta_2)

. The sine function has the property that

\sin(x) = \sin(180° - x)

. Therefore, if

2\theta_1 \neq 2\theta_2

, we must have

2\theta_1 = 180° - 2\theta_2

. Dividing by 2 gives

\theta_1 = 90° - \theta_2

, or

\theta_1 + \theta_2 = 90°

. The two angles are complementary. Distractor C is only true for the maximum range, where

\theta_1 = \theta_2 = 45°

. The other distractors represent common misconceptions.

Question 6

A train starts from rest and accelerates at $a_1$ for a time $t_1$ . It then immediately decelerates at $a_2$ until it comes to rest. What is the total distance travelled by the train?

$\frac{1}{2} (a_1 + a_2) t_1^2$
$\frac{1}{2} a_1 t_1^2 (1 + \frac{a_1}{a_2})$ (correct answer)
$\frac{1}{2} (a_1 - a_2) t_1^2$
$\frac{1}{2} \frac{a_1 a_2}{a_1 + a_2} t_1^2$

Explanation: Stage 1 (acceleration): Distance

s_1 = \frac{1}{2}a_1 t_1^2

. The maximum velocity reached is

v_{max} = a_1 t_1

. Stage 2 (deceleration): The train decelerates from

v_{max}

to 0. Let the time taken be

t_2

0 = v_{max} - a_2 t_2 \Rightarrow t_2 = \frac{v_{max}}{a_2} = \frac{a_1 t_1}{a_2}

. The distance covered is

s_2 = v_{max}t_2 - \frac{1}{2}a_2 t_2^2 = (a_1 t_1)(\frac{a_1 t_1}{a_2}) - \frac{1}{2}a_2(\frac{a_1 t_1}{a_2})^2 = \frac{a_1^2 t_1^2}{a_2} - \frac{1}{2}\frac{a_1^2 t_1^2}{a_2} = \frac{1}{2}\frac{a_1^2 t_1^2}{a_2}

. Total distance

s = s_1 + s_2 = \frac{1}{2}a_1 t_1^2 + \frac{1}{2}\frac{a_1^2 t_1^2}{a_2} = \frac{1}{2}a_1 t_1^2 (1 + \frac{a_1}{a_2})

. This is a multi-step problem requiring derivation, making it challenging. The distractors represent plausible but incorrect algebraic combinations.

Question 7

Two stones, P and Q, are thrown from the same point at the same time with the same initial speed. Stone P is thrown vertically upwards, and stone Q is thrown at 30° to the horizontal. Neglecting air resistance, what is true about the stones just after launch?

They have the same initial acceleration. (correct answer)
They have the same initial vertical velocity component.
They will reach the same maximum height.
They have the same initial horizontal velocity component.

Explanation: Once the stones are in the air, the only force acting on both of them is gravity (neglecting air resistance). Therefore, both stones have the same constant downward acceleration,

g

. This is true for any projectile motion near the Earth's surface. Distractor B is incorrect; P has initial vertical velocity

u

, while Q has

u \sin(30°) = 0.5u

. Distractor C is incorrect; since their initial vertical velocities are different, they will reach different maximum heights. Distractor D is incorrect; P has zero horizontal velocity, while Q has a non-zero horizontal velocity.

Question 8

A car travelling at speed $v$ can brake to a stop in a distance $d$ . Assuming the same constant braking force, what is the stopping distance if the car is initially travelling at speed $3v$ ?

$d$
$3d$
$6d$
$9d$ (correct answer)

Explanation: We can use the kinematic equation

v_f^2 = v_i^2 + 2as

. Here, the final speed

v_f = 0

, initial speed is

v_i

, acceleration is

a

(which is negative), and stopping distance is

s=d

. So,

0 = v_i^2 + 2ad \Rightarrow d = \frac{-v_i^2}{2a}

. This shows that the stopping distance

d

is proportional to the square of the initial speed (

d \propto v_i^2

), since the acceleration

a

is constant (due to constant braking force). If the initial speed is tripled to

3v_i

, the new stopping distance

d'

will be

d' \propto (3v_i)^2 = 9v_i^2

. Therefore, the new stopping distance is

9d

. Distractor B is a common error from assuming a linear relationship between speed and stopping distance.

Question 9

Car A starts from rest at $t=0$ and moves with a constant acceleration of $2.0$ m s⁻². Car B passes car A at $t=0$ , moving in the same direction with a constant velocity of $8.0$ m s⁻¹. At what time after $t=0$ does car A overtake car B?

2.0 s
4.0 s
8.0 s (correct answer)
16.0 s

Explanation: Let the starting point be

x=0

. The position of car A is given by

x_A = u_A t + \frac{1}{2}a_A t^2 = 0 + \frac{1}{2}(2.0)t^2 = t^2

. The position of car B is given by

x_B = v_B t = 8.0t

. Car A overtakes car B when their positions are the same, i.e.,

x_A = x_B

t^2 = 8.0t

. Since we are looking for a time after

t=0

, we can divide by

t

t = 8.0

s. Distractor B is the time when car A's speed reaches car B's speed (v = at => 8 = 2t => t=4). At this point, car B is still ahead, but car A is starting to close the gap faster.

Question 10

An elastic ball is dropped from rest onto a hard horizontal surface and bounces once. The effects of air resistance and the duration of contact with the surface are negligible. Taking the upward direction as positive, which statement best describes the acceleration-time graph of the motion?

A constant horizontal line at a negative value. (correct answer)
A constant negative value, with a large positive spike during the bounce.
A constant positive value throughout the motion.
A line with a constant negative slope.

Explanation: While the ball is in the air (both falling and rising), the only force acting on it is gravity (since air resistance is negligible). Therefore, its acceleration is constant and equal to

-g

. The question states the duration of contact is negligible, implying we consider only the time the ball is in free-fall. During this entire time, the acceleration is constant. The large upward acceleration during the bounce itself occurs over a negligible time and would not appear on a graph representing the overall motion. Distractor B correctly identifies the acceleration during contact, but the question focuses on the motion where contact time is negligible, so the dominant feature is constant acceleration due to gravity. Distractor C has the wrong sign. Distractor D describes a changing acceleration.

Question 11

A particle undergoes one complete revolution in a circle of radius $R$ at a constant speed $v$ . What are the magnitudes of the average velocity and the average acceleration for this complete revolution?

Average velocity is $v$ ; average acceleration is 0.
Average velocity is 0; average acceleration is $\frac{v^2}{R}$ .
Average velocity is 0; average acceleration is 0. (correct answer)
Average velocity is $v$ ; average acceleration is $\frac{v^2}{R}$ .

Explanation: Average velocity is defined as total displacement divided by total time,

\vec{v}_{avg} = \frac{\Delta\vec{x}}{\Delta t}

. After one complete revolution, the particle returns to its starting point, so the total displacement

\Delta\vec{x}

is zero. Thus, the average velocity is zero. Average acceleration is defined as the change in velocity divided by total time,

\vec{a}_{avg} = \frac{\Delta\vec{v}}{\Delta t}

. After one complete revolution, the particle is at the same point with the same instantaneous velocity (both magnitude and direction) as it had at the start. Therefore, the change in velocity

\Delta\vec{v}

is zero, and the average acceleration is also zero. Distractor B incorrectly confuses the constant magnitude of the instantaneous centripetal acceleration with the average acceleration.

Question 12

A projectile is launched from ground level with kinetic energy $E$ at an angle of 60° to the horizontal. Neglecting air resistance, what is the kinetic energy of the projectile at the highest point of its trajectory?

0
$0.25 E$ (correct answer)
$0.50 E$
$0.75 E$

Explanation: The initial kinetic energy is

E = \frac{1}{2}mu^2

, where

u

is the initial speed. At the highest point of the trajectory, the vertical component of velocity is zero, but the horizontal component remains constant:

v_x = u \cos\theta

. The speed at the highest point is

v = u \cos(60°)

. Since

\cos(60°) = 0.5

, the speed is

v = 0.5u

. The kinetic energy at this point is

E_{top} = \frac{1}{2}mv^2 = \frac{1}{2}m(0.5u)^2 = \frac{1}{2}mu^2(0.5)^2 = E \times 0.25

. Distractor A incorrectly assumes the speed is zero at the top. Distractor C forgets to square the cosine term. Distractor D uses

\sin^2(60°)

, which relates to the potential energy gained.

Question 13

A golf ball is hit at an angle above the horizontal. Air resistance has a significant effect on the flight of the ball. Which statement best describes the trajectory of the ball compared to the ideal parabolic path in a vacuum?

The maximum height achieved is lower and the horizontal range is shorter. (correct answer)
The trajectory remains a symmetric parabola, but with a reduced range.
The time of flight is longer because air resistance slows the ball down.
The angle of impact with the ground is less steep than the launch angle.

Explanation: Air resistance is a dissipative force that removes energy from the system. This means the ball will not travel as high or as far as it would in a vacuum. Therefore, both the maximum height and the horizontal range are reduced. Distractor B is incorrect because air resistance makes the trajectory asymmetric; the descent is typically steeper than the ascent. Distractor C is incorrect; because the maximum height is lower and the ball is slowed in both horizontal and vertical directions, the time of flight is shorter. Distractor D is incorrect; the descent is steeper, meaning the angle of impact is greater than the launch angle.

Question 14

A ball is dropped from rest and falls freely for 3.0 s before hitting the ground. From what height was it dropped, and what was its velocity just before impact? (Use g = 10 m/s²)

Height = 90 m; Impact velocity = 60 m/s, considering the acceleration increases during the fall
Height = 90 m; Impact velocity = 30 m/s, accounting for the average velocity during the fall
Height = 45 m; Impact velocity = 60 m/s, using energy conservation instead of kinematic equations
Height = 45 m; Impact velocity = 30 m/s, using basic kinematic equations for free fall (correct answer)

Explanation: When you encounter free fall problems, you're dealing with constant acceleration due to gravity, making this perfect for kinematic equations. The key insight is that acceleration remains constant at g = 10 m/s² throughout the entire fall. For height, use the kinematic equation

h = v_0t + \frac{1}{2}gt^2

. Since the ball starts from rest,

v_0 = 0

, so:

h = \frac{1}{2}(10)(3.0)^2 = \frac{1}{2}(10)(9) = 45 \text{ m}

For impact velocity, use

v = v_0 + gt

. Again with

v_0 = 0

v = (10)(3.0) = 30 \text{ m/s}

Option A incorrectly suggests acceleration increases during fall, but gravity provides constant acceleration near Earth's surface. It also miscalculates both values. Option B gets the height wrong by doubling it to 90 m, possibly confusing the kinematic equation or incorrectly applying average velocity concepts. The impact velocity is also wrong. Option C correctly identifies that energy methods would work (

mgh = \frac{1}{2}mv^2

), but the height calculation is wrong, leading to an incorrect velocity of 60 m/s. The correct answer is D: height = 45 m and impact velocity = 30 m/s using standard kinematic equations. Study tip: For IB Physics free fall problems, always start with the three kinematic equations and identify which variables you know. Remember that near Earth's surface, g remains constant, and initial velocity is zero when objects are "dropped from rest."

Question 15

Two particles start at the same point and travel in the same direction. Particle A has constant velocity 8 m/s. Particle B starts from rest with constant acceleration 1.6 m/s². After how much time will particle B be 10 m ahead of particle A?

t = 5.0 s, when particle B's velocity first equals particle A's velocity
t = 10.0 s, when both particles have traveled equal distances from the start
t = 12.5 s, calculated by solving the position difference equation (correct answer)
t = 15.0 s, representing twice the catch-up time for the additional separation

Explanation: Position equations:

x_A = 8t

and

x_B = \frac{1}{2}(1.6)t^2 = 0.8t^2

. For B to be 10 m ahead:

x_B - x_A = 10

, so

0.8t^2 - 8t = 10

, giving

0.8t^2 - 8t - 10 = 0

. Multiplying by 1.25:

t^2 - 10t - 12.5 = 0

. Using the quadratic formula:

t = \frac{10 ± \sqrt{100 + 50}}{2} = \frac{10 ± \sqrt{150}}{2} = \frac{10 ± 5\sqrt{6}}{2}

. Taking the positive root:

t ≈ 12.25

s ≈ 12.5 s. Choice A is when velocities are equal (

v_B = v_A

1.6t = 8

, so

t = 5

s). Choice B is when they meet (

x_A = x_B

Question 16

An object is dropped from rest from a height $H$ . It takes time $t_1$ to fall the first $H/2$ and time $t_2$ to fall the second $H/2$ . What is the ratio $\frac{t_2}{t_1}$ ?

1
$\sqrt{2}$
$\sqrt{2} - 1$ (correct answer)
$\frac{1}{\sqrt{2}}$

Explanation: For the first half of the fall:

s = ut + \frac{1}{2}at^2 \Rightarrow \frac{H}{2} = 0 + \frac{1}{2}gt_1^2 \Rightarrow t_1 = \sqrt{\frac{H}{g}}

. For the total fall:

H = 0 + \frac{1}{2}g(t_1+t_2)^2 \Rightarrow t_1+t_2 = \sqrt{\frac{2H}{g}} = \sqrt{2}\sqrt{\frac{H}{g}} = \sqrt{2} \, t_1

. Therefore,

t_2 = \sqrt{2} \, t_1 - t_1 = (\sqrt{2}-1)t_1

. The ratio is

\frac{t_2}{t_1} = \sqrt{2}-1

. Distractor A assumes linear motion. Distractor B is the ratio of total time to

t_1

. Distractor D is a common algebraic error.

Question 17

An object's position $x$ as a function of time $t$ is given by $x = At^3 - Bt$ , where A and B are positive constants. At what time is the object's acceleration equal to zero?

0 (correct answer)
$\sqrt{B/A}$
$\sqrt{B/(3A)}$
At no time, as acceleration is constant.

Explanation: To find acceleration, we must differentiate the position function twice with respect to time. Velocity

v = \frac{dx}{dt} = 3At^2 - B

. Acceleration

a = \frac{dv}{dt} = 6At

. We want to find the time

t

when

a=0

. Setting

6At = 0

, and since

A

is a positive constant, we find that

t=0

. Distractor C gives the time when the velocity is zero. Distractor B is an algebraic error. Distractor D incorrectly assumes the acceleration is constant.

Question 18

A small sphere is released from rest in a tall column of viscous liquid. Which statement correctly describes the net force on the sphere as it falls?

The net force is initially zero and increases to a constant maximum value.
The net force is constant throughout the motion.
The net force is initially at its maximum value and decreases, approaching zero. (correct answer)
The net force is always zero.

Explanation: The net force on the sphere is

F_{net} = F_{gravity} - F_{buoyancy} - F_{drag}

. At

t=0

, the sphere is at rest, so the drag force (which depends on speed) is zero. The net force is

F_{gravity} - F_{buoyancy}

, which is its maximum value. As the sphere accelerates and its speed increases, the drag force increases, opposing the motion. This causes the net downward force to decrease. If the column is tall enough, the sphere will reach terminal velocity, where the drag force becomes large enough that the net force approaches zero. Distractor A has the logic reversed. Distractor B would imply constant acceleration, which is false. Distractor D would imply no motion.

Question 19

A car accelerates uniformly from rest. It travels 40 m in the third second of its motion. What is its acceleration? The first second is from t=0 to t=1 s.

8 m s⁻²
10 m s⁻²
16 m s⁻² (correct answer)
20 m s⁻²

Explanation: The third second of motion is the interval between

t=2

s and

t=3

s. The distance travelled in time

t

s(t) = ut + \frac{1}{2}at^2

. Since

u=0

s(t) = \frac{1}{2}at^2

. Distance at

t=3

s(3) = \frac{1}{2}a(3^2) = 4.5a

. Distance at

t=2

s(2) = \frac{1}{2}a(2^2) = 2a

. The distance travelled in the third second is

s(3) - s(2) = 4.5a - 2a = 2.5a

. We are given this distance is 40 m. So,

2.5a = 40 \Rightarrow a = \frac{40}{2.5} = 16

m s⁻². Distractors arise from incorrect interpretation of 'third second' or calculation errors, e.g., setting

s(3) = 40

, which gives

a \approx 8.9

, or confusing average velocity.

Question 20

An object moves with a constant, non-zero acceleration. Which statement about the object's motion must be true?

Its speed is always increasing.
Its velocity changes at a constant rate. (correct answer)
Its direction of motion does not change.
Its displacement in equal time intervals is the same.

Explanation: The definition of constant acceleration is that the velocity changes by the same amount in each unit of time, i.e., it changes at a constant rate. Distractor A is false; if acceleration is opposite to the initial velocity, the object will slow down (speed decreases). Distractor C is false; in projectile motion, the acceleration is constant (due to gravity), but the direction of motion continuously changes. Distractor D is false; for accelerated motion, the distance covered in successive time intervals increases (if speeding up) or decreases (if slowing down), as shown by the

t^2

term in

s = ut + \frac{1}{2}at^2