Kinematics
Unit 1 of AP Physics 1, worth 10–15% of the exam. 13 questions below, each with the working. Every answer was checked by a second pass before it was published.
Scalars and vectors, displacement, velocity and acceleration, representations of motion, reference frames, motion in two dimensions.
How this unit is tested
Kinematics questions on the AP exam almost always start by asking you to identify what kind of quantity you're dealing with (scalar or vector) and what kind of motion is happening (constant velocity, constant acceleration, or projectile motion). Before touching an equation, decide on a positive direction and stick with it for the whole problem — most sign errors come from switching conventions halfway through.
Next, identify whether acceleration is constant. If it is, the four kinematic equations apply directly; if it isn't (e.g., a graph with a curved v-t line, or a verbal description of changing acceleration), you must work from the definitions (average velocity = Δx/Δt, average acceleration = Δv/Δt) or from areas and slopes on graphs instead. The AP exam frequently tests graph literacy: position-time slope gives velocity, velocity-time slope gives acceleration, and area under a velocity-time or acceleration-time graph gives displacement or change in velocity respectively.
For two-dimensional motion, the single most important habit is splitting the problem into independent horizontal and vertical components at the very start. Horizontal velocity is constant (no horizontal force in projectile motion), while vertical motion is uniformly accelerated by gravity. Solve each axis with the one-dimensional tools you already have, then recombine using vector addition (magnitude and direction, or components) only at the end if the question asks for a resultant vector.
Finally, always check your answer against the physical picture: does the sign of velocity match the direction of motion described, is the time positive, does the trajectory make sense? The exam rewards students who can reason qualitatively about motion, not just plug into formulas.
What you have to know
Average velocity and acceleration
Average velocity is v_avg = Δx/Δt (displacement over time interval). Average acceleration is a_avg = Δv/Δt (change in velocity over time interval).
Kinematic equations (constant acceleration only)
v = v0 + a t; x = x0 + v0 t + (1/2) a t^2; v^2 = v0^2 + 2 a (x − x0); x − x0 = (1/2)(v0 + v) t.
Graph relationships
On a position-time graph, the slope equals velocity. On a velocity-time graph, the slope equals acceleration and the area under the curve equals displacement. On an acceleration-time graph, the area under the curve equals the change in velocity.
Independence of perpendicular motion
In two-dimensional motion, horizontal and vertical components of velocity and acceleration are independent of each other; each can be analyzed with one-dimensional kinematics separately.
Relative velocity
The velocity of object A relative to frame C equals the velocity of A relative to B plus the velocity of B relative to C: $\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$.
13 practice questions
-
A runner completes one full lap around a 400 m circular track, returning exactly to the starting point. What are the runner's total distance traveled and total displacement?
Show the answer
Answer. Distance = 400 m; displacement = 0 m.
Distance is the total path length traveled, which is the full circumference (400 m). Displacement is the vector from start to end position; since the runner returns to the same point, the displacement is zero regardless of the path taken. -
An object's position is x(t) = 3t^2 − 2t (in meters, t in seconds). Find its instantaneous velocity at t = 2 s.
Show the answer
Answer. 10 m/s
Instantaneous velocity is the derivative of position with respect to time: v(t) = 6t − 2. At t = 2 s, v = 6(2) − 2 = 10 m/s. -
A velocity-time graph shows an object's velocity increasing linearly from 0 to 8 m/s over 4 seconds, then remaining constant at 8 m/s for the next 2 seconds. What is the total displacement over the 6 seconds?
Show the answer
Answer. 32 m
Displacement equals the area under the v-t graph. The first section is a triangle: (1/2)(4 s)(8 m/s) = 16 m. The second section is a rectangle: (2 s)(8 m/s) = 16 m. Total = 16 + 16 = 32 m. -
A car starts from rest and accelerates uniformly at 3 m/s^2. How far does it travel in the first 5 seconds?
Show the answer
Answer. 37.5 m
Use x = x0 + v0t + (1/2)at^2 with v0 = 0: x = 0 + 0 + (1/2)(3)(5^2) = (1/2)(3)(25) = 37.5 m. -
A ball is dropped from rest from a height of 20 m. Ignoring air resistance and using g = 10 m/s^2, what is its speed just before hitting the ground?
Show the answer
Answer. 20 m/s
Use v^2 = v0^2 + 2aΔy with v0 = 0, a = 10 m/s^2, Δy = 20 m: v^2 = 2(10)(20) = 400, so v = 20 m/s. -
An object moves with velocity in the +x direction while its acceleration points in the −x direction. What is happening to the object's speed?
Show the answer
Answer. The object is slowing down (decelerating).
When acceleration is opposite in direction to velocity, it acts to reduce the magnitude of velocity, so the object's speed decreases even though it may still be moving in the +x direction. -
A position-time graph shows a straight line with negative slope. What does this indicate about the object's motion?
Show the answer
Answer. The object moves with constant velocity in the negative direction.
A straight line on a position-time graph means constant slope, i.e., constant velocity. A negative slope means the velocity is negative, so the object moves steadily in the negative direction. -
An acceleration-time graph shows a constant acceleration of 4 m/s^2 for 3 seconds, starting from an initial velocity of 2 m/s. What is the velocity at the end of the 3 seconds?
Show the answer
Answer. 14 m/s
The area under an a-t graph gives the change in velocity: Δv = (4 m/s^2)(3 s) = 12 m/s. Adding to the initial velocity: v = 2 + 12 = 14 m/s. -
A projectile is launched horizontally at 15 m/s from the top of a 45 m tall cliff. Using g = 10 m/s^2, how far from the base of the cliff does it land?
Show the answer
Answer. about 45 m
First find the fall time from the vertical motion: 45 = (1/2)(10)t^2 → t^2 = 9 → t = 3 s. Horizontal distance is constant velocity: x = v_x t = 15(3) = 45 m. -
A projectile is launched at an angle above the horizontal. At the very top of its trajectory (maximum height), what are its vertical and horizontal velocity components?
Show the answer
Answer. Vertical velocity is zero; horizontal velocity is unchanged from launch (nonzero).
At maximum height the vertical velocity momentarily equals zero before reversing direction, while the horizontal velocity remains constant throughout the flight since there is no horizontal acceleration. -
A boat can travel at 4 m/s in still water and heads straight across a river that flows at 3 m/s. What is the boat's speed relative to the ground?
Show the answer
Answer. 5 m/s
The boat's velocity relative to ground is the vector sum of its velocity relative to water (4 m/s, across) and the water's velocity relative to ground (3 m/s, downstream). Since these are perpendicular, magnitude = √(4^2+3^2) = √25 = 5 m/s. -
Two cars travel in the same direction on a highway: Car A at 30 m/s and Car B at 22 m/s. What is the velocity of Car A relative to Car B?
Show the answer
Answer. 8 m/s in the direction of travel (Car A moves away from Car B at 8 m/s).
Relative velocity of A with respect to B is v_A − v_B = 30 − 22 = 8 m/s in the same direction both cars are moving, meaning Car A pulls ahead of Car B at 8 m/s. -
An object undergoes non-uniform acceleration described only by a curved (non-straight) line on a velocity-time graph. Which method correctly finds the average acceleration over a time interval?
Show the answer
Answer. Compute Δv/Δt using the velocity values at the start and end of the interval (the slope of the secant line), not a kinematic equation.
When acceleration is not constant, the kinematic equations (which assume constant a) do not apply. Average acceleration is still defined as the change in velocity divided by the change in time, corresponding to the slope of the secant line connecting the two points on the v-t curve.
What people get wrong
- Treating speed and velocity as interchangeable. Speed is the magnitude of velocity and is always positive; velocity carries direction and can be negative. Always ask whether the question wants magnitude only or a signed/directional answer.
- Applying the four kinematic equations when acceleration is not constant (e.g., during a curved velocity-time graph). Instead, use the definitions of average velocity/acceleration or read slopes and areas directly off the graph.
- Assuming horizontal velocity changes during projectile motion. In the absence of air resistance, a_x = 0, so horizontal velocity is constant throughout the flight — only vertical velocity changes.
- Forgetting to decompose the initial velocity into components when a projectile is launched at an angle, and mistakenly plugging the full speed into the vertical equations. Always split into v0x and v0y first.
- Confusing the slope of a position-time graph with the area under a velocity-time graph. Slope of x-t gives velocity; area under v-t gives displacement — mixing these up leads to using the wrong graph feature.
- Picking inconsistent positive directions mid-problem, which causes sign errors in acceleration or final velocity. Choose one positive direction at the start and keep every vector's sign consistent with it.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Physics 1. Free, and no account needed to start.