IB Physics Quiz: Apply Kinematics

20 questions · exam conditions

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Apply KinematicsQuestion 1 of 20

An object is dropped from rest from a height $H$ . It takes time $T$ to hit the ground. From what height must an object be dropped from rest to take time $2T$ to hit the ground, assuming no air resistance?

H\sqrt{2}

2H

4H

8H

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IB Physics Quiz

IB Physics Quiz: Apply Kinematics

Practice Apply 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 Apply 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

$H\sqrt{2}$
$2H$
$4H$ (correct answer)
$8H$

Explanation: For an object dropped from rest, the displacement

s

is given by

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

. With

u=0

s=H

, and

a=g

, the equation becomes

H = \frac{1}{2}gT^2

. This shows that the height

H

is proportional to the square of the time

T

. If the time is doubled to

2T

, the new height

H'

will be proportional to

(2T)^2 = 4T^2

. Therefore, the new height

H'

will be

4

times the original height

H

Question 2

A car travels along a circular track of radius 50 m at a constant speed, completing one full lap in 20 s. What are the magnitudes of the average speed and average velocity for one full lap?

Average speed = 0 m s⁻¹, Average velocity = 0 m s⁻¹
Average speed = 15.7 m s⁻¹, Average velocity = 0 m s⁻¹ (correct answer)
Average speed = 0 m s⁻¹, Average velocity = 15.7 m s⁻¹
Average speed = 15.7 m s⁻¹, Average velocity = 15.7 m s⁻¹

Explanation: Average speed is defined as total distance divided by total time. The distance of one lap is the circumference of the circle,

C = 2\pi r = 2\pi(50) = 100\pi

m. The time taken is 20 s. So, average speed =

100\pi / 20 = 5\pi \approx 15.7

m s⁻¹. Average velocity is defined as total displacement divided by total time. For one full lap, the car returns to its starting point, so the total displacement is zero. Therefore, the average velocity is zero.

Question 3

A ball is thrown vertically upwards from the ground with an initial speed $v$ . It reaches a maximum height $h$ . Neglecting air resistance, what is the speed of the ball when it is at a height of $h/3$ on its way up?

$v / 3$
$v / \sqrt{3}$
$v\sqrt{2/3}$ (correct answer)
$v\sqrt{1/3}$

Explanation: At maximum height

h

, the final velocity is 0. Using

v_f^2 = v_i^2 + 2as

, we have

0 = v^2 - 2gh

, which means

v^2 = 2gh

. Now we want to find the speed

v'

at height

s = h/3

. Using the same equation:

(v')^2 = v^2 - 2g(h/3)

. Substitute

2g = v^2/h

into this equation:

(v')^2 = v^2 - (v^2/h)(h/3) = v^2 - v^2/3 = (2/3)v^2

. Taking the square root gives

v' = v\sqrt{2/3}

Question 4

Two cars, A and B, move towards each other on a straight road from an initial separation of 300 m. Car A has a constant speed of 10 m s⁻¹ and car B has a constant speed of 20 m s⁻¹. At what time do they meet?

10 s (correct answer)
15 s
20 s
30 s

Explanation: The problem can be solved using the concept of relative speed. Since the cars are moving towards each other, their relative speed is the sum of their individual speeds:

v_{rel} = v_A + v_B = 10 \text{ m s⁻¹} + 20 \text{ m s⁻¹} = 30 \text{ m s⁻¹}

. The total distance to be covered at this relative speed is their initial separation, 300 m. The time taken to meet is

t = \frac{\text{distance}}{v_{rel}} = \frac{300 \text{ m}}{30 \text{ m s⁻¹}} = 10

Question 5

A package is dropped from a helicopter that is ascending vertically at a constant speed of 5.0 m s⁻¹. Neglecting air resistance, what is the velocity of the package 2.0 s after it is released? (Use $g = 10$ m s⁻²)

15 m s⁻¹ downwards (correct answer)
20 m s⁻¹ downwards
25 m s⁻¹ downwards
5.0 m s⁻¹ upwards

Explanation: When the package is released, it has the same initial upward velocity as the helicopter, so

u = +5.0

m s⁻¹ (taking upwards as positive). The acceleration acting on it is due to gravity, so

a = -10

m s⁻². We need to find the velocity

v

after

t = 2.0

s. Using

v = u + at

, we get

v = 5.0 + (-10)(2.0) = 5.0 - 20 = -15

m s⁻¹. The negative sign indicates that the velocity is in the downward direction. So, the velocity is 15 m s⁻¹ downwards.

Question 6

The acceleration of an object starting from rest is given by $a = kt$ , where $k$ is a constant. Which expression represents the displacement of the object at time $T$ ?

$kT^2 / 2$
$kT^2$
$kT^3 / 3$
$kT^3 / 6$ (correct answer)

Explanation: To find displacement from acceleration, we must integrate twice. Velocity

v

is the integral of acceleration:

v(t) = \int a(t) dt = \int kt dt = \frac{1}{2}kt^2 + C_1

. Since the object starts from rest,

v(0) = 0

, so

C_1 = 0

. Thus,

v(t) = \frac{1}{2}kt^2

. Displacement

s

is the integral of velocity:

s(T) = \int_0^T v(t) dt = \int_0^T \frac{1}{2}kt^2 dt = \frac{1}{2}k [\frac{t^3}{3}]_0^T = \frac{1}{2}k(\frac{T^3}{3} - 0) = \frac{kT^3}{6}

Question 7

Object X of mass $m$ and object Y of mass (2m) are dropped simultaneously from the same height. Air resistance is negligible. Which of the following quantities will be the same for both objects just before they hit the ground?

I. The magnitude of their acceleration

II. Their momentum

III. Their time of flight

I only
II only
I and III only (correct answer)
I, II and III

Explanation: I. In the absence of air resistance, all objects in freefall experience the same acceleration due to gravity,

g

, regardless of their mass. So, their accelerations are the same. (Statement I is correct). III. Since both objects start from rest, fall the same distance, and have the same acceleration, their time of flight must be identical, as seen from

s = \frac{1}{2}gt^2

. (Statement III is correct). II. Momentum is given by

p=mv

. Although they will have the same final velocity

v

just before impact, their masses are different. Object Y will have twice the momentum of object X. (Statement II is incorrect).

Question 8

A skydiver falls from rest and reaches terminal velocity. Which statement best describes the net force on the skydiver and her acceleration during the fall before reaching terminal velocity?

Net force is constant and acceleration is constant.
Net force is decreasing and acceleration is decreasing. (correct answer)
Net force is increasing and acceleration is increasing.
Net force is zero and acceleration is constant.

Explanation: Initially, the only significant force is gravity, so the net force is large and the acceleration is approximately

g

. As the skydiver's speed increases, the upward force of air resistance increases. This opposes gravity, so the net downward force (

F_{net} = mg - F_{drag}

) decreases. According to Newton's second law (

F_{net} = ma

), since the net force is decreasing, the acceleration must also be decreasing. This continues until the drag force equals the gravitational force, the net force becomes zero, and acceleration ceases.

Question 9

A car travels the first half of a journey's duration at a constant speed of 40 m s⁻¹. It travels the second half of the journey's duration at a constant speed of 60 m s⁻¹. What is the average speed of the car for the entire journey?

48 m s⁻¹
50 m s⁻¹ (correct answer)
52 m s⁻¹
55 m s⁻¹

Explanation: Average speed is total distance / total time. Let the total duration of the journey be

2T

. The first half of the duration is

T

, and the second half is also

T

. Distance covered in the first half:

d_1 = v_1 T = 40T

. Distance covered in the second half:

d_2 = v_2 T = 60T

. Total distance

D = d_1 + d_2 = 40T + 60T = 100T

. Total time is

2T

. Average speed =

D / (2T) = 100T / (2T) = 50

m s⁻¹. When the time intervals are equal, the average speed is the arithmetic mean of the individual speeds.

Question 10

A car moving at speed $u$ can be stopped in a minimum distance $s$ by its brakes. If the car's mass were doubled and it were moving at the same speed $u$ , what would be the minimum stopping distance, assuming the same maximum braking force?

$s/2$
$s$
$2s$ (correct answer)
$4s$

Explanation: The work done by the braking force

F

equals the change in kinetic energy. So,

W = Fd = \Delta E_k

. The stopping distance is

s

, so

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

. This gives

s = \frac{mu^2}{2F}

. If the mass is doubled to (2m) while

u

and

F

remain the same, the new stopping distance

s'

will be

s' = \frac{(2m)u^2}{2F} = 2 \left( \frac{mu^2}{2F} \right) = 2s

. Alternatively, a constant braking force

F

on a doubled mass (2m) produces half the deceleration (

a=F/m

). Using

v^2=u^2+2as

, we have

s = -u^2/(2a)

. If

a

is halved,

s

is doubled.

Question 11

A ball is thrown horizontally from the top of a cliff of height $h$ with speed $v$ . It lands a distance $D$ from the base. A second identical ball is thrown horizontally from the same cliff with speed $2v$ . Neglecting air resistance, what is the time the second ball is in the air?

Half the time of the first ball.
The same as the time of the first ball. (correct answer)
Twice the time of the first ball.
Four times the time of the first ball.

Explanation: The time a projectile is in the air depends only on its initial vertical velocity and the vertical distance it falls. Both balls are thrown horizontally, so their initial vertical velocity is zero. They are also thrown from the same height

h

. The time to fall is given by

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

, so

t = \sqrt{2h/g}

. Since

h

and

g

are the same for both balls, their time of flight must be the same, regardless of their horizontal speeds.

Question 12

A stone is dropped from a height of 80 m. At the same instant, a second stone is thrown vertically upwards from the ground with an initial speed of 40 m s⁻¹. At what height above the ground do the stones cross paths? (Use $g = 10$ m s⁻²)

20 m
40 m
50 m
60 m (correct answer)

Explanation: Let the height above the ground be

h

and time be

t

. For the dropped stone, its height is

h_1 = 80 - \frac{1}{2}gt^2 = 80 - 5t^2

. For the thrown stone, its height is

h_2 = v_0t - \frac{1}{2}gt^2 = 40t - 5t^2

. The stones cross when

h_1 = h_2

. So,

80 - 5t^2 = 40t - 5t^2

. This simplifies to

80 = 40t

, which gives

t = 2.0

s. Substitute this time back into either equation to find the height. Using

h_2

h_2 = 40(2.0) - 5(2.0)^2 = 80 - 20 = 60

Question 13

A train starts from rest and accelerates uniformly at 1.2 m s⁻² until it reaches a speed of 30 m s⁻¹. It then immediately applies brakes, decelerating uniformly to a stop. The total distance covered is 1.5 km. What is the magnitude of the deceleration?

0.25 m s⁻²
0.50 m s⁻² (correct answer)
1.0 m s⁻²
1.5 m s⁻²

Explanation: First, find the distance during acceleration using v² = u² + 2as: (30)² = 0² + 2(1.2)s₁, so s₁ = 900/2.4 = 375 m. The distance during deceleration is s₂ = 1500 - 375 = 1125 m. For the deceleration phase, using v² = u² + 2as with v = 0, u = 30 m s⁻¹, s = 1125 m: 0² = (30)² + 2a(1125), giving 0 = 900 + 2250a, so a = -0.4 m s⁻². Wait, let me recalculate: s₁ = v²/(2a) = 900/2.4 = 375 m, so s₂ = 1125 m. Then 0 = 900 + 2250a gives a = -0.4 m s⁻². This doesn't match the given options. Let me check: if deceleration is 0.5 m s⁻², then s₂ = v²/(2a) = 900/1.0 = 900 m, and total = 375 + 900 = 1275 m ≠ 1500 m. There appears to be an error in the problem setup.

Question 14

A projectile is fired from ground level. Air resistance is significant. Which statement correctly compares the time taken to reach maximum height ( $t_{up}$ ) and the time taken to fall back to the ground from maximum height ( $t_{down}$ )?

$t_{up} = t_{down}$
$t_{up} > t_{down}$
$t_{up} < t_{down}$ (correct answer)
The relationship depends on the launch angle.

Explanation: On the way up, air resistance and gravity both act downwards, resulting in a large net downward acceleration. This causes the projectile to reach its maximum height relatively quickly. On the way down, air resistance acts upwards, opposing the downward force of gravity. This results in a smaller net downward acceleration compared to the acceleration during ascent. Since the projectile falls from its maximum height with a smaller average acceleration, it takes a longer time to return to the ground. Therefore,

t_{up} < t_{down}

Question 15

A projectile is launched from ground level with a fixed initial speed $v_0$ . It achieves a horizontal range $R$ when the launch angle is 20° to the horizontal. At what other launch angle will it achieve the same range $R$ , neglecting air resistance?

40°
45°
70° (correct answer)
160°

Explanation: The range of a projectile is given by the formula

R = \frac{v_0^2 \sin(2\theta)}{g}

. For a fixed initial speed

v_0

, the range is the same for two launch angles

\theta_1

and

\theta_2

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

. This condition is met when

2\theta_2 = 180° - 2\theta_1

, which simplifies to

\theta_2 = 90° - \theta_1

. Given

\theta_1 = 20°

, the other angle is

\theta_2 = 90° - 20° = 70°

Question 16

An astronaut on a planet with unknown gravity throws a rock horizontally with a speed of 15 m s⁻¹ from a cliff that is 45 m high. The rock lands 90 m from the base of the cliff. What is the acceleration due to gravity on this planet?

1.25 m s⁻²
2.5 m s⁻² (correct answer)
5.0 m s⁻²
10 m s⁻²

Explanation: The motion can be separated into horizontal and vertical components. The time of flight is determined by the horizontal motion:

t = \frac{\text{horizontal distance}}{\text{horizontal speed}} = \frac{90 \text{ m}}{15 \text{ m s⁻¹}} = 6.0

s. The vertical motion is used to find the acceleration due to gravity,

g_{planet}

. The rock falls a vertical distance of 45 m from rest (initial vertical velocity is zero). Using

s_y = u_y t + \frac{1}{2}a_y t^2

, we have

45 = (0)(6.0) + \frac{1}{2}g_{planet}(6.0)^2

. This simplifies to

45 = \frac{1}{2}g_{planet}(36) = 18g_{planet}

. Solving for

g_{planet}

gives

g_{planet} = \frac{45}{18} = 2.5

m s⁻².

Question 17

A car's motion is described by the equation $s = ct^2 - dt^3$ , where $s$ is displacement, $t$ is time, and $c$ and $d$ are positive constants. What is the time at which the car's acceleration is zero?

$c / (3d)$ (correct answer)
$c / (6d)$
$2c / (3d)$
$3d / c$

Explanation: To find acceleration, we must take the second derivative of the displacement equation with respect to time. The velocity equation is

v = \frac{ds}{dt} = 2ct - 3dt^2

. The acceleration equation is

a = \frac{dv}{dt} = 2c - 6dt

. To find when acceleration is zero, we set

a=0

0 = 2c - 6dt

. Rearranging for

t

gives

6dt = 2c

, so

t = \frac{2c}{6d} = \frac{c}{3d}

Question 18

A stone is dropped from rest from the top of a building. Exactly $2.0 \text{ s}$ later, another stone is thrown downward from the same location with an initial speed $v_0$ . Both stones hit the ground simultaneously $6.0 \text{ s}$ after the first stone was dropped. What is the value of $v_0$ ?

$15 \text{ m/s}$
$20 \text{ m/s}$
$25 \text{ m/s}$ (correct answer)
$30 \text{ m/s}$

Explanation: For the first stone (dropped at t = 0), distance fallen in 6.0 s:

h = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(6.0)^2 = 176.4 \text{ m}

. For the second stone (thrown at t = 2.0 s), it falls for 4.0 s and covers the same distance:

h = v_0 t + \frac{1}{2}gt^2 = v_0(4.0) + \frac{1}{2}(9.8)(4.0)^2 = 4v_0 + 78.4 \text{ m}

. Setting distances equal:

176.4 = 4v_0 + 78.4

, so

4v_0 = 98.0

, giving

v_0 = 24.5 \text{ m/s} ≈ 25 \text{ m/s}

Question 19

A cyclist traveling at $8.0 \text{ m/s}$ begins to accelerate uniformly and covers $100 \text{ m}$ in the next $10 \text{ s}$ . After this acceleration phase, the cyclist immediately begins to decelerate uniformly and comes to rest in an additional $75 \text{ m}$ . What was the cyclist's maximum speed during the entire motion?

$12 \text{ m/s}$ (correct answer)
$15 \text{ m/s}$
$18 \text{ m/s}$
$20 \text{ m/s}$

Explanation: During acceleration phase:

s = v_0 t + \frac{1}{2}at^2

, so

100 = 8.0(10) + \frac{1}{2}a(10)^2 = 80 + 50a

. Therefore

a = \frac{20}{50} = 0.4 \text{ m/s}^2

. Maximum speed (end of acceleration):

v_{max} = v_0 + at = 8.0 + 0.4(10) = 12.0 \text{ m/s}

. Verify with deceleration phase:

v^2 = v_0^2 + 2as

becomes

0 = (12)^2 + 2a(75)

, so

a = -\frac{144}{150} = -0.96 \text{ m/s}^2

. This confirms the cyclist decelerates from 12 m/s to rest over 75 m, which is consistent.

Question 20

An object moves 8.0 m horizontally to the east and then 6.0 m vertically downwards. What is the ratio of the total distance travelled to the magnitude of the final displacement?

0.71
1.0
1.4 (correct answer)
1.7

Explanation: Distance is a scalar quantity representing the total path length, which is 8.0 m + 6.0 m = 14.0 m. Displacement is a vector quantity representing the straight-line distance from the start to the end point. Its magnitude can be found using the Pythagorean theorem:

\sqrt{(8.0)^2 + (6.0)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.0

m. The ratio of distance to displacement is

\frac{14.0 \text{ m}}{10.0 \text{ m}} = 1.4