A car starts from rest and speeds up at a constant 3.0 m/s^2 for 6.0 s. How far does it travel in that time?
Keep practicing
Now make your own
Upload your prep book, slides or notes, or just tell Bo what you want to learn. You do not need a document to start.
About this test
This is a free 15-question sampler in the style of the AP Physics 1 exam. It touches all eight units, from kinematics and forces through work and energy, momentum, rotation, oscillations and fluids, and mixes multiple choice, select all that apply and fill in the blank. Answer all 15 questions, then check your score and read the explanation for every question.
All 15 questions at a glance
Prefer to read before you play? Every question in the sampler is listed below. Answers and explanations stay collapsed until you open them.
1.A car starts from rest and speeds up at a constant 3.0 m/s^2 for 6.0 s. How far does it travel in that time?
- A. 54 m
- B. 18 m
- C. 36 m
- D. 108 m
Answer: A. 54 m
Starting from rest, the distance is d = v0 t + (1/2) a t^2 = 0 + (1/2)(3.0)(6.0)^2 = 54 m. The 18 is a t, which is the final speed in m/s and not a distance. The 108 comes from leaving out the one half.
2.On a velocity-versus-time graph, an object's line is straight and rises from 2 m/s at t = 0 to 10 m/s at t = 4 s. Which statements are correct?
Select all that apply.
- A. The acceleration is 2 m/s^2
- B. The displacement during the 4 s is 24 m
- C. The acceleration is 2.5 m/s^2
- D. The object is speeding up
- E. The displacement during the 4 s is 40 m
Answer: A. The acceleration is 2 m/s^2, B. The displacement during the 4 s is 24 m, D. The object is speeding up
The slope of a velocity-time graph is the acceleration: (10 - 2) / 4 = 2 m/s^2. The area under the graph is the displacement. With a straight line the average velocity is (2 + 10) / 2 = 6 m/s, so the displacement is 6 x 4 = 24 m. Speed and acceleration point the same way, so the object is speeding up. The 2.5 m/s^2 divides the final velocity by the time, and 40 m multiplies the final velocity by the time, which treats the object as if it had moved at 10 m/s the whole way.
3.A person pushes a box across a rough horizontal floor with a constant horizontal force, and the box moves at a constant velocity. Which statement is correct?
- A. The push is greater than the force of kinetic friction
- B. The net force on the box points in the direction of motion
- C. The normal force on the box equals the push
- D. The push equals the force of kinetic friction in size
Answer: D. The push equals the force of kinetic friction in size
A constant velocity means zero acceleration, so by Newton's second law the net force is zero. Horizontally, the push and the kinetic friction must be equal in size and opposite in direction. Vertically, the normal force balances the weight, not the push. A box does not need a net force to keep moving, only to speed up or slow down.
4.A 60 kg person stands on a scale in an elevator that is accelerating upward at 2.0 m/s^2. Using g = 9.8 m/s^2, what does the scale read?
- A. 468 N
- B. 588 N
- C. 708 N
- D. 120 N
Answer: C. 708 N
Two forces act on the person: the scale pushes up with the normal force N, and gravity pulls down with mg. Taking up as positive, N - mg = ma, so N = m(g + a) = 60 x (9.8 + 2.0) = 708 N. The 588 N is the person's weight, which is what the scale reads at rest or at constant velocity, and 468 N is what it would read if the elevator were accelerating downward at 2.0 m/s^2.
5.Complete the sentence.
In a head-on collision between a heavy truck and a small car, the force the truck exerts on the car is _____ the force the car exerts on the truck, but the car's acceleration is _____ the truck's.
Options for blank 1: greater than, equal in size to, less than, unrelated to
Options for blank 2: greater than, equal to, smaller than, in the same direction as
Answer: equal in size to, greater than
Newton's third law says the two forces form an interaction pair: equal in size, opposite in direction, acting on different objects. Acceleration is a = F / m, and the same size force on a much smaller mass gives a much larger acceleration. That is why the car's motion changes far more than the truck's even though neither one pushes harder.
6.A 2.0 kg ball is released from rest 5.0 m above the floor. Ignoring air resistance, about how fast is it moving just before it hits the floor?
- A. 4.9 m/s
- B. 9.9 m/s
- C. 14 m/s
- D. 49 m/s
Answer: B. 9.9 m/s
Energy is conserved, so the potential energy lost equals the kinetic energy gained: mgh = (1/2)mv^2. The mass cancels, leaving v = sqrt(2gh) = sqrt(2 x 9.8 x 5.0) = sqrt(98), about 9.9 m/s. A heavier ball would hit the floor at the same speed.
7.A 10 kg crate slides across a rough floor with an initial speed of 6.0 m/s. The only horizontal force on it is a constant 30 N force of kinetic friction. How far does it slide before it stops?
- A. 0.6 m
- B. 1.2 m
- C. 3.0 m
- D. 6.0 m
Answer: D. 6.0 m
The work done by friction equals the change in kinetic energy. The crate starts with (1/2)(10)(6.0)^2 = 180 J and ends with zero, so friction must remove 180 J. Friction removes 30 N x d, so d = 180 / 30 = 6.0 m. You can check it with kinematics: a = 30 / 10 = 3.0 m/s^2, and d = v^2 / (2a) = 36 / 6 = 6.0 m.
8.A motor lifts a 50 kg crate straight up at a constant speed of 0.40 m/s. Using g = 9.8 m/s^2, what power does the motor deliver to the crate?
- A. 20 W
- B. 196 W
- C. 490 W
- D. 1,960 W
Answer: B. 196 W
At constant speed the net force is zero, so the motor's upward force equals the weight: F = mg = 50 x 9.8 = 490 N. Power is force times velocity in the direction of the force, so P = Fv = 490 x 0.40 = 196 W. The 490 is the force in newtons and not a power, and 20 W comes from multiplying mass by speed and skipping g.
9.A 2.0 kg cart moving at 3.0 m/s collides with a 1.0 kg cart at rest, and the two stick together. What is their speed right after the collision?
- A. 1.0 m/s
- B. 1.5 m/s
- C. 2.0 m/s
- D. 3.0 m/s
Answer: C. 2.0 m/s
No external horizontal force acts on the system, so momentum is conserved: (2.0)(3.0) + (1.0)(0) = (2.0 + 1.0)v, which gives v = 2.0 m/s. Kinetic energy is not conserved in a collision where objects stick together. It drops from 9.0 J to 6.0 J, and the missing 3.0 J goes into deformation and thermal energy.
10.A 0.50 kg ball moving at 4.0 m/s hits a wall and bounces straight back at 4.0 m/s. The contact lasts 0.020 s. Which statements are correct?
Select all that apply.
- A. The ball's momentum is unchanged because its speed is unchanged
- B. The ball's momentum changes by 4.0 kg m/s
- C. The average force of the wall on the ball is 200 N
- D. The impulse on the ball is 2.0 N s
- E. The ball's kinetic energy is unchanged
Answer: B. The ball's momentum changes by 4.0 kg m/s, C. The average force of the wall on the ball is 200 N, E. The ball's kinetic energy is unchanged
Momentum is a vector, so reversing direction changes it even at the same speed: the change is m(v - v0) = 0.50 x (-4.0 - 4.0) = -4.0 kg m/s, a size of 4.0. That change is the impulse, and the average force is impulse over time: 4.0 / 0.020 = 200 N. Kinetic energy depends on speed squared and has no direction, so it is unchanged. The 2.0 N s is what you get by forgetting that the velocity reversed.
11.A 40 N force is applied to a wrench 0.25 m from the bolt. The force makes an angle of 30 degrees with the handle. What is the size of the torque about the bolt?
- A. 5.0 N m
- B. 2.5 N m
- C. 8.7 N m
- D. 10 N m
Answer: A. 5.0 N m
Torque is the lever arm times the force times the sine of the angle between them: tau = rF sin(theta) = (0.25)(40) sin 30 degrees = 10 x 0.5 = 5.0 N m. Only the part of the force perpendicular to the handle turns the bolt. The 10 N m would need the force at 90 degrees, and 8.7 N m comes from using cosine instead of sine.
12.Complete the sentence.
A spinning ice skater pulls her arms in. Her rotational inertia decreases, so her angular speed _____, and because no external torque acts on her, her angular momentum _____.
Options for blank 1: increases, decreases, stays the same, drops to zero
Options for blank 2: increases, decreases, stays the same, reverses direction
Answer: increases, stays the same
With no net external torque, angular momentum L = I omega is conserved. When the moment of inertia I decreases, omega must increase so that the product stays the same. Her rotational kinetic energy goes up, which is fine: her muscles do work as they pull her arms in.
13.Complete the sentence.
For a mass on a spring, quadrupling the mass multiplies the period by _____. Changing the amplitude _____ the period.
Options for blank 1: 0.5, 2, 4, 16
Options for blank 2: does not change, doubles, halves, quadruples
Answer: 2, does not change
The period of a mass-spring oscillator is T = 2 pi sqrt(m / k). The period grows with the square root of the mass, so four times the mass gives sqrt(4) = 2 times the period. The amplitude does not appear in the formula. A larger stretch means a larger restoring force, so the mass moves faster and covers the longer path in the same time.
14.A diver is 10 m below the surface of a freshwater lake (density 1,000 kg/m^3). Using g = 9.8 m/s^2, what is the gauge pressure at that depth?
- A. 9.8 kPa
- B. 98 kPa
- C. 196 kPa
- D. 980 kPa
Answer: B. 98 kPa
Gauge pressure from the water alone is P = rho g h = (1,000)(9.8)(10) = 98,000 Pa, or 98 kPa. Absolute pressure would add atmospheric pressure of about 101 kPa on top, which roughly doubles it, but gauge pressure leaves the atmosphere out.
15.A solid block floats in water with exactly half of its volume below the surface. Which statements are correct?
Select all that apply.
- A. The buoyant force equals the block's weight
- B. The block's density is half the density of water
- C. The buoyant force is greater than the block's weight
- D. The block's density equals the density of water
- E. The buoyant force equals the weight of the water the block displaces
Answer: A. The buoyant force equals the block's weight, B. The block's density is half the density of water, E. The buoyant force equals the weight of the water the block displaces
A floating object is in equilibrium, so the buoyant force equals its weight. Archimedes' principle says the buoyant force always equals the weight of the fluid displaced. Setting the weight of the block equal to the weight of the displaced water gives rho_block x V = rho_water x (V / 2), so the block's density is half that of water. If the buoyant force were greater than the weight, the block would rise out of the water. If the density matched the water, the block would float fully submerged.
What the AP Physics 1 exam looks like and how it is weighted
AP Physics 1 is the algebra-based, first-year college physics course in the AP program, scored 1 to 5. According to the College Board, the format changed starting with the May 2027 exam. Section I is 42 multiple-choice questions in 85 minutes, worth 50 percent of the score. Section II is 4 free-response questions in 95 minutes, worth the other 50 percent, one each of mathematical routines, translation between representations, experimental design and analysis, and qualitative and quantitative translation. The exam is hybrid digital: you answer the multiple-choice questions and read the free-response questions in the Bluebook app, and you handwrite the free-response answers in a paper booklet. Calculators are permitted and reference information is provided. Older prep books list a different number of questions and different timing, so check the College Board page before you plan your pacing.
The course has eight units, and the College Board publishes how much of the multiple-choice section each one gets. Force and translational dynamics and work, energy and power are the heaviest at 18 to 23 percent each. Kinematics, linear momentum, torque and rotational dynamics, and fluids each carry 10 to 15 percent, and energy and momentum of rotating systems and oscillations carry 5 to 8 percent each. The unit list was reorganized, so a book or a video series written for an older version of the course may not match the Course and Exam Description. Use the current one as your reference.
A 5 is the top score, but what a score is worth is up to each college. Some give credit for a 3, many ask for a 4 or a 5, and some give none for AP Physics 1. Look up the policy at the schools you care about before you decide how hard to push.
How to use this sampler
The 15 questions are spread across all eight units, with the most questions on the units the exam weights most. They are meant to show you which unit to open next rather than to predict a score. Each one asks for the reasoning, not only the number: why a box moving at constant speed still has friction acting on it, what stays the same when a skater pulls her arms in, why the period of a spring does not care how far you pull it.
Take it in one sitting with a calculator but without notes, using g = 9.8 m/s^2 where a question needs it. Then read the explanation for every question, including the ones you got right. Treat a miss as a topic. If you missed the elevator question and the friction question, the gap is not two formulas, it is the habit of drawing a free-body diagram and writing Newton's second law for each direction. Put units on every number when you check your work, since a missing unit is an easy way to lose a point on the free-response section.
The questions are original. They are written to the published course outline and do not use any released exam question. This sampler practices the multiple-choice style of reasoning. It does not practice the four long free-response questions, so pair it with past free-response questions from the College Board for that half of the exam.
How StudyPDF builds AP Physics 1 practice tests from your class notes
Fifteen questions can show you which units are shaky. They cannot get you ready on their own, and a generic question bank rarely matches the labs and problem sets your teacher actually assigned.
StudyPDF works from your material instead. Upload your textbook chapters, your teacher's slides, a lab write-up or your problem sets, press New material, and StudyPDF builds practice tests from those exact pages. Every question is grounded in your material and every explanation cites where the answer comes from, so you can go straight back to the page when a step does not make sense.
You do not need to upload anything to start. Name a topic instead, for example torque and angular momentum or conservation of energy, and Bo sets it up with its key concepts so you can build quizzes and flashcards on it. Your results track which concepts you keep missing, so the next practice test can target them. Starting is free.
Written by the StudyPDF team. Last updated August 19, 2026.
Good to know
Is this AP Physics 1 practice test free?
Yes. All 15 questions, the score screen and every explanation are free, and the quiz runs in your browser. A free StudyPDF account is needed to download the PDF or to build more practice tests from your own notes.
Who wrote these questions?
The StudyPDF team wrote every question and explanation, following the eight units of the published AP Physics 1 course outline. They are original items, not real or released exam questions, and StudyPDF is not affiliated with the College Board, which does not endorse this page.
How is this different from the real AP Physics 1 exam?
The real multiple-choice section is 42 questions in 85 minutes starting with the May 2027 exam, and the exam adds 4 free-response questions in 95 minutes. This sampler is 15 questions with no time limit, and it also uses select all that apply and fill in the blank, because judging each statement separately is a good check of how well you know a unit. It does not practice the free-response section.
Can I get practice tests built from my own class materials?
Yes. Upload your textbook chapters, slides, lab write-ups or problem sets to StudyPDF, and you can build practice tests from those exact pages, with explanations tied to your sources. If you have nothing to upload, name a topic or a unit instead.