Kinematics practice problems AP Physics C: Mechanics, Unit 1
Position, velocity and acceleration as derivatives and integrals, motion graphs, and projectile motion. 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 for kinematics
| v = dx/dt, a = dv/dt | velocity and acceleration as derivatives |
|---|---|
| x = x0 + ∫ v dt | position from velocity |
| v = v0 + at | constant acceleration only |
| x = x0 + v0t + ½at2 | constant acceleration only |
Use g = 9.8 m/s², as on the AP reference sheet. You can type answers like 4.9, 1/3 or 3.6e2. The full sheet is in our AP Physics C equation sheet.
Warm-up questions
Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.
Hint
Differentiate x(t).
Worked solution
- v = dx/dt = 6t − 2.
- At t = 2: v = 10 m/s.
Hint
Integrate a(t) and use the initial condition.
Worked solution
- v = ∫ 6t dt = 3t2 + C, and v(0) = 2 gives C = 2.
- v(3) = 27 + 2 = 29 m/s.
Hint
Which component of velocity changes, and which does not?
Worked solution
- The vertical velocity is zero at the top, but the horizontal velocity is unchanged.
- So the velocity is horizontal, while the acceleration is still g downwards: C.
Exam-style questions
Questions at the level of the AP Physics C: Mechanics exam. The ones marked Challenge combine more than one idea.
Hint
Displacement is the integral of velocity.
Worked solution
- Δx = ∫02 (4t3 − 3t2) dt = [t4 − t3]02.
- Δx = 16 − 8 = 8 m.
Hint
Only the vertical component matters for height.
Worked solution
- Vertical component: 20 sin 30° = 10 m/s.
- At the top vy = 0: h = vy2/(2g) = 100/19.6.
- h = 5.10 m.
Hint
Integrate twice, using both initial conditions.
Worked solution
- v = t2 + 1 (using v(0) = 1).
- x = t3/3 + t (using x(0) = 0).
- x(3) = 9 + 3 = 12 m.
Hint
At rest means v = 0.
Worked solution
- v = 3t2 − 12t + 9 = 3(t − 1)(t − 3).
- v = 0 at t = 1 s and t = 3 s: B.
Hint
Find the time to fall 45 m from the vertical motion; the horizontal velocity stays 12 m/s.
Worked solution
- Vertically: 45 = ½(9.8)t2, so t = 3.03 s.
- Horizontally: x = 12 × 3.03 = 36.4 m.
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 15 minutes
A particle moves along the x-axis with position x(t) = t3 − 6t2 + 9t, in metres, for 0 ≤ t ≤ 4 s.
- (a) Find expressions for the velocity v(t) and acceleration a(t).
- (b) Find the times when the particle is at rest, and its position at each.
- (c) Find the total distance travelled from t = 0 to t = 4 s.
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)
v = 3t2 − 12t + 9; a = 6t − 12.
Part (b)
v = 3(t − 1)(t − 3) = 0 at t = 1 s (x = 4 m) and t = 3 s (x = 0).
Part (c)
x(0) = 0, x(1) = 4, x(3) = 0, x(4) = 4. Distance = 4 + 4 + 4 = 12 m.
Your mark: 0 of 6
Common mistakes in kinematics
- Using constant-acceleration equations when the acceleration depends on time. Integrate instead.
- Confusing displacement with distance: split the motion where v changes sign.
- Forgetting the constant of integration, or not using the initial condition to find it.
- Using the full launch speed instead of its vertical component for height and time of flight.
Keep going
Revise the ideas behind these questions:
Then try real exam questions: AP Physics C: Mechanics past papers, with official scoring guidelines. · All AP Physics C: Mechanics units