AP Physics 1 / Topic

Kinematics

Unit 1 describes motion without asking why it happens: position, velocity, and acceleration in one and two dimensions, motion graphs, free fall, projectiles, and relative motion.

44 practice questions3 FRQs4 skill areas

What the exam asks

  • Translate among position-time, velocity-time, and acceleration-time graphs (slopes and areas)
  • Apply the constant-acceleration equations to 1D motion and free fall
  • Break projectile motion into independent horizontal and vertical components
  • Reason about relative velocity between reference frames
  • Distinguish average from instantaneous quantities and speed from velocity

Key formulas and rules

Constant acceleration

v = v0 + at; x = x0 + v0t + ½at²; v² = v0² + 2aΔx

Graph rules

slope of x-t is velocity; slope of v-t is acceleration; area under v-t is displacement

Free fall

a = g ≈ 10 m/s² downward, independent of mass

Projectiles

horizontal: constant vx; vertical: free fall; time links the two

At the peak

vy = 0 but a is still g downward

Question bank breakdown

Multiple choice

44

Free response

3

Difficulty mix

11 easy · 25 medium · 8 hard

Skills covered

Displacement, Velocity, and AccelerationVectors and Motion in Two DimensionsReference Frames and Relative MotionRepresenting Motion

Every question in the bank comes with a per-choice explanation, so you learn why each wrong answer is wrong — not just the key.

Sample question

From the free tier of the ScoreMint AP Physics 1 bank — try it, then check the answer.

A car starts from rest and accelerates uniformly at $2 \text{ m/s}^2$ for $10 \text{ s}$. It then travels at constant velocity for another $10 \text{ s}$. What is the total distance traveled by the car during the entire $20 \text{ s}$?

  • A. $100 \text{ m}$
  • B. $200 \text{ m}$
  • C. $300 \text{ m}$
  • D. $400 \text{ m}$
Show answer and explanation

Answer: C. This problem requires breaking the motion into two phases. During uniform acceleration from rest: $d_1 = \frac{1}{2}(2)(10)^2 = 100$ m, reaching $v = 20$ m/s. During constant velocity: $d_2 = 20 \times 10 = 200$ m. Total distance = $300$ m. A common mistake is applying the acceleration equations over the entire time interval, but the kinematic equations for constant acceleration only apply within each phase where acceleration is constant.

Study tip

For any projectile question, write two columns (x and y) and solve for time first — time is the only quantity the two directions share.