Kinematics practice problems AP Physics 1, Unit 1

Written by STEM-Tutor's tutors · Updated October 2026 · 10 questions and 1 free-response question

Motion in one and two dimensions: displacement, velocity and acceleration, motion graphs, free fall and projectiles. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page

Key equations
Warm-up questions
Exam-style questions
Free-response practice
Common mistakes
Keep going

Key equations for kinematics

v_x = v_x0 + a_xt	velocity after time t at constant acceleration
x = x₀ + v_x0t + ½a_xt²	position after time t
v_x² = v_x0² + 2a_x(x − x₀)	when time is not given or not asked for
v_avg = Δx / Δt	average velocity

All questions use g = 9.8 m/s², as on the AP Physics 1 equation sheet. A calculator is allowed, as in the exam. The full sheet is in our AP Physics 1 equation sheet.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up

A cyclist speeds up uniformly from 3.0 m/s to 9.0 m/s in 4.0 s. What is the cyclist's acceleration?

Hint

Acceleration is the change in velocity divided by the time taken.

Worked solution

Use a = Δv / Δt.
Δv = 9.0 − 3.0 = 6.0 m/s.
a = 6.0 / 4.0 = 1.5 m/s².

Q2Warm-up

A ball is thrown straight up. At the very top of its path, which statement is true?

Hint

Gravity does not switch off at the top. What is the only force acting on the ball?

Worked solution

The only force on the ball is gravity, the whole time it is in the air.
So its acceleration is g = 9.8 m/s² downward everywhere, including at the top.
At the top, the velocity is changing from upward to downward, so it is zero for an instant.
The answer is B. A zero acceleration would mean the ball stays at the top.

Q3Warm-up

A car travelling at 20 m/s brakes uniformly and stops in 50 m. What is the magnitude of its acceleration?

Hint

No time is given or asked for, so pick the equation without t.

Worked solution

Use v² = v₀² + 2aΔx with v = 0.
0 = 20² + 2a(50), so 0 = 400 + 100a.
a = −4.0 m/s². The minus sign means the acceleration is opposite to the motion.
Magnitude: 4.0 m/s².

Q4Warm-up

On a velocity-time graph, how do you find the displacement of an object?

Hint

Think about the units: velocity (m/s) multiplied by time (s).

Worked solution

Velocity × time gives displacement, and on a graph that product is an area.
The slope of a velocity-time graph is the acceleration, not the displacement.
The answer is B: displacement is the area between the graph and the time axis. Area below the axis counts as negative displacement.

Exam-style questions

Questions at the level of the AP Physics 1 exam. The ones marked Challenge combine more than one idea.

Q5Exam-style

A stone is dropped from rest from a bridge and hits the water 2.5 s later. Ignoring air resistance, how high is the bridge above the water?

Hint

Dropped from rest means v₀ = 0. Which equation links distance, time and acceleration?

Worked solution

Take downward as positive, so a = g = 9.8 m/s² and v₀ = 0.
Δy = v₀t + ½gt² = 0 + ½(9.8)(2.5)².
Δy = 4.9 × 6.25 = 30.6 m.

Q6Exam-style

A ball rolls off a horizontal table 1.2 m high at 3.0 m/s. How far from the foot of the table does it land?

Hint

Treat the vertical and horizontal motions separately. The vertical motion tells you the time in the air.

Worked solution

Vertically the ball starts with zero velocity and falls 1.2 m: 1.2 = ½(9.8)t².
t = √(2 × 1.2 / 9.8) = 0.495 s.
Horizontally the velocity stays at 3.0 m/s, so Δx = 3.0 × 0.495.
Δx = 1.48 m.

Q7Exam-style

A projectile is launched from level ground at 20 m/s, 30° above the horizontal. How long is it in the air?

Hint

Find the vertical component of the launch velocity first. The time up equals the time down.

Worked solution

Vertical component: v_y0 = 20 sin 30° = 10 m/s.
Time to the top, where v_y = 0: t = 10 / 9.8 = 1.02 s.
The flight is symmetric on level ground, so the total time is 2 × 1.02 = 2.04 s, answer B.
1.02 s is only the time to the top, a common trap.

Q8Exam-style

The position-time graph of a cart is a straight line sloping downward. Which describes the cart's motion?

Hint

On a position-time graph, the slope is the velocity. What does a constant, negative slope mean?

Worked solution

The slope of a position-time graph is the velocity.
A straight line has a constant slope, so the velocity is constant and the acceleration is zero.
The slope is negative, so the cart moves in the negative direction.
The answer is B.

Q9Challenge

Car A starts from rest and accelerates at 2.0 m/s² at the moment car B passes it at a constant 12 m/s. How long does car A take to catch car B?

Hint

Write an expression for each car's position against time, then set them equal.

Worked solution

Car A: x_A = ½(2.0)t² = t².
Car B: x_B = 12t.
Set them equal: t² = 12t, so t(t − 12) = 0.
t = 0 is the start, so A catches B at t = 12 s.

Q10Challenge

A ball is thrown straight up at 15 m/s from 2.0 m above the ground. How long is it in the air before it hits the ground?

Hint

Take up as positive and the launch point as y = 0. Where is the ground?

Worked solution

Take up as positive, so a = −9.8 m/s², and the ground is at y = −2.0 m.
−2.0 = 15t − 4.9t², so 4.9t² − 15t − 2.0 = 0.
Quadratic formula: t = [15 + √(225 + 39.2)] / 9.8 = (15 + 16.25) / 9.8.
The positive root is t = 3.19 s.

Free-response practice

One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.

Mathematical routines style · 6 points · about 20 minutes

A student throws a ball horizontally at 8.0 m/s from the top of a building of height H = 20 m. Air resistance is negligible.

A projectile launched horizontally: constant horizontal velocity, constant downward acceleration g. Time links the two directions.

(a) Derive an expression for the time t the ball takes to reach the ground, in terms of H and physical constants.
(b) Calculate the horizontal distance from the base of the building to where the ball lands.
(c) A second ball is thrown horizontally from the same point at 16 m/s. Compare its time in the air and its landing distance with the first ball's. Justify your answers.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Vertically the ball starts from rest: H = ½gt², so t = √(2H/g).

1 point for: Uses a correct vertical equation with zero initial vertical velocity, such as H = ½gt²
1 point for: Correct expression t = √(2H/g)

Part (b)

t = √(2 × 20 / 9.8) = 2.02 s. Horizontal velocity is constant, so d = 8.0 × 2.02 = 16.2 m.

1 point for: Uses d = v_xt with the time from part (a)
1 point for: Correct answer, 16 m to 16.2 m, with units

Part (c)

The time is the same, because the vertical motion (start at rest vertically, fall H with acceleration g) does not depend on the horizontal velocity. The distance doubles, because d = v_xt and v_x doubles while t stays the same.

1 point for: Same time, justified by the independence of vertical and horizontal motion
1 point for: Distance doubles, justified using d = v_xt with the same t

Your mark: 0 of 6

Common mistakes in kinematics

Mixing up the time to the top with the total flight time of a projectile.
Using 9.8 m/s² for the horizontal acceleration of a projectile: horizontally the acceleration is zero.
Forgetting that velocity and acceleration have signs. Choose a positive direction at the start and keep to it.
Reading a position-time graph as if it were a velocity-time graph.

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Physics 1 past papers, with official scoring guidelines. · All AP Physics 1 units