Master Kinematics with AP Physics 1 review games.
This unit covers displacement and velocity, acceleration, projectile motion and kinematic equations — essential concepts for AP Physics 1. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers displacement and velocity, acceleration, projectile motion and kinematic equations — essential concepts for AP Physics 1. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Displacement and Velocity
Displacement is a vector quantity representing the change in position (Δx = x_f - x_i), not total distance traveled. Average velocity is displacement divided by elapsed time (v_avg = Δx/Δt), while instantaneous velocity is the slope of a position-time graph at a point. The AP exam frequently tests whether students can distinguish scalar distance/speed from vector displacement/velocity.
Key Points
- Displacement can be negative; distance is always non-negative
- Average velocity = Δx/Δt; average speed = total distance/Δt — these are not the same
- On a position-time graph, slope = velocity; a horizontal line means v = 0
- An object can have zero displacement but nonzero distance (e.g., round trip)
A car drives 60 m east, then 20 m west in 8 seconds. What is its average velocity? What is its average speed?
Displacement is 60 − 20 = 40 m east, so average velocity = 40 m / 8 s = 5 m/s east. Total distance is 60 + 20 = 80 m, so average speed = 80 m / 8 s = 10 m/s. This distinguishes the two quantities — average velocity depends only on start and end positions, not the path taken.
2 Acceleration
Acceleration is the rate of change of velocity (a = Δv/Δt) and is a vector; its sign indicates direction, not whether the object is speeding up or slowing down. An object slows down when velocity and acceleration point in opposite directions. The AP exam heavily tests interpreting velocity-time graphs, where slope = acceleration and area under the curve = displacement.
Key Points
- Negative acceleration does not mean slowing down — it means acceleration points in the negative direction
- On a v-t graph: slope = acceleration, area between curve and time axis = displacement
- Constant (uniform) acceleration produces a straight line on a v-t graph
- Zero acceleration means constant velocity, not necessarily zero velocity
A velocity-time graph shows a straight line from v = 10 m/s at t = 0 to v = −2 m/s at t = 6 s. Find the acceleration and the displacement over this interval.
Acceleration = Δv/Δt = (−2 − 10)/6 = −2 m/s², meaning the object decelerates while moving in the positive direction, stops momentarily, then moves in the negative direction. Displacement equals the area under the v-t graph: a triangle above the axis (area = ½ × 5 × 10 = 25 m) plus a triangle below (area = ½ × 1 × 2 = −1 m), giving net displacement = 24 m. Note that total distance traveled (26 m) differs from displacement.
3 Projectile Motion
Projectile motion involves independent horizontal and vertical components: horizontal velocity is constant (a_x = 0), while vertical motion has constant downward acceleration g = 9.8 m/s². The two components share only time as a linking variable. The AP exam tests setting up two separate sets of kinematic equations and correctly identifying initial velocity components using trigonometry.
Key Points
- Horizontal: v_x = v_0 cosθ (constant); Vertical: v_y0 = v_0 sinθ (changes due to gravity)
- At maximum height, v_y = 0; horizontal velocity is never zero (for standard launch)
- Time of flight is determined entirely by vertical motion
- Range is maximized at 45° for launch and landing at the same height
A ball is launched from the ground at 20 m/s at 30° above horizontal. How long is it in the air, and how far does it travel horizontally? (g = 10 m/s²)
The vertical initial velocity is v_y0 = 20 sin30° = 10 m/s. Using Δy = v_y0 t − ½gt² with Δy = 0 (lands at same height): 0 = 10t − 5t², giving t = 2 s. Horizontal distance = v_x × t = (20 cos30°)(2) = (17.3)(2) ≈ 34.6 m. The key step is solving vertical motion first to find time, then using that time in the horizontal equation.
4 Kinematic Equations
The four kinematic equations apply only under constant acceleration and relate five variables: displacement (Δx), initial velocity (v_0), final velocity (v), acceleration (a), and time (t). For each problem, identify which variable is unknown and which is not given (the 'missing variable'), then select the equation that excludes the missing variable. These equations are provided on the AP formula sheet but students must know how to apply them correctly.
Key Points
- v = v_0 + at (missing: Δx)
- Δx = v_0 t + ½at² (missing: v)
- v² = v_0² + 2aΔx (missing: t)
- Δx = ½(v_0 + v)t (missing: a) — use when acceleration is unknown but not needed
A car traveling at 30 m/s brakes with a constant deceleration of 6 m/s². How far does it travel before stopping?
Known: v_0 = 30 m/s, v = 0 m/s (stops), a = −6 m/s²; unknown: Δx; missing variable: t. Select v² = v_0² + 2aΔx. Substituting: 0 = 900 + 2(−6)Δx → Δx = 900/12 = 75 m. The sign of acceleration must be negative (opposing motion), and setting v = 0 is the key physical condition that defines 'stopping.'
Questions, answered.
What is Kinematics?
Kinematics is Unit 1 of AP Physics 1, covering displacement and velocity, acceleration, projectile motion and kinematic equations.
How to study for AP Physics 1 Unit 1?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 30+ review questions across 5 different game modes.