Kinematics worksheet answers for AP Physics 1 (free PDF download)

Problem 1 Distance and displacement

1. Distance = 40 + 25 = 65 m.
2. Δx = +40 − 25 = +15 m, that is 15 m east.
3. Average speed = 65 / 50 = 1.3 m/s. Average velocity = +15 / 50 = +0.30 m/s (east).

Problem 2 Scalar or vector

Scalars (no direction): distance, speed, time.
Vectors (direction needed): displacement, velocity, acceleration.
In one dimension the sign of a vector tells its direction relative to the chosen positive direction: a negative value points the opposite way.

Problem 3 Train between two stations

1. Δx = 14.0 − 2.0 = 12.0 km = 12 000 m.
2. 8.0 min = 8.0 × 60 = 480 s.
3. v = 12 000 / 480 = 25 m/s; × 3.6 gives 90 km/h.

Problem 4 Motorbike pulling away

1. a = Δv/Δt = (23 − 5.0) / 6.0 = 3.0 m/s².
2. v = 23 + 3.0 × 4.0 = 35 m/s.

Problem 5 Reading a position-time graph

1. Slope: 0–4 s: 8/4 = +2.0 m/s; 4–8 s: 0 m/s (at rest); 8–14 s: (−4 − 8)/6 = −2.0 m/s.
2. Distance = 8 + 0 + 12 = 20 m; displacement = −4 − 0 = −4 m.
3. Average velocity = −4/14 = −0.29 m/s; average speed = 20/14 = 1.4 m/s.

Problem 6 Reading a velocity-time graph

1. Stage 1: 8/2 = +4.0 m/s²; stage 2: 0; stage 3: (−4 − 8)/4 = −3.0 m/s².
2. Area = ½ × 2 × 8 + 4 × 8 = 8 + 32 = 40 m.
3. From 8 m/s with −3.0 m/s²: t = 6 + 8/3 = 8.7 s.
4. From 6 s to 8.67 s: ½ × 2.67 × 8 = 10.7 m; from 8.67 s to 10 s: ½ × 1.33 × (−4) = −2.7 m. Position = 40 + 10.7 − 2.7 = 48 m.

Problem 7 Skateboard on a ramp

1. Δx = v₀t + ½at² with v₀ = 0: it contains Δx, t and a, and the unknown is a.
2. a = 2Δx/t² = 2 × 6.0 / 2.0² = 3.0 m/s².
3. v = at = 3.0 × 2.0 = 6.0 m/s.

Problem 8 Metro train braking

1. v² = v₀² + 2aΔx gives a = (8.0² − 20²) / (2 × 168) = (64 − 400)/336 = −1.0 m/s².
2. t = (v − v₀)/a = (8.0 − 20)/(−1.0) = 12 s.

Problem 9 Falling coconut

1. t = √(2h/g) = √(24.0/9.81) = √2.446 = 1.56 s.
2. v = gt = 9.81 × 1.564 = 15.3 m/s; or v = √(2gh) = √(2 × 9.81 × 12.0) = √235.4 = 15.3 m/s.

Problem 10 Velocity table of a thrown ball

1. v = 9.0 − 9.81t: +4.1 m/s (0.50 s), −0.81 m/s (1.00 s), −5.7 m/s (1.50 s), −10.6 m/s (2.00 s).
2. v = 0: t = 9.0/9.81 = 0.92 s.
3. h = v₀²/(2g) = 81/19.62 = 4.1 m.

Problem 11 Signs of velocity and acceleration

1. Same signs: speeding up, moving right.
2. Opposite signs: slowing down, moving right.
3. Same signs: speeding up, moving left.
4. Opposite signs: slowing down, moving left.

Problem 12 Matatu sprint

1. 108 / 3.6 = 30 m/s.
2. a = 30 / 7.5 = 4.0 m/s².
3. Δx = ½(v₀ + v)t = ½ × 30 × 7.5 = 112.5 m ≈ 1.1 × 10² m.

Problem 13 Spot the sign error

1. She chose down as positive but then used the sign for an upward-positive axis. In addition, the acceleration points down, which is the positive direction she chose.
2. a = +9.81 m/s², so v = 0 + 9.81 × 2.0 = +19.6 m/s, which means 19.6 m/s downward.

Problem 14 Scooter journey

1. v = 1.5 × 8.0 = 12 m/s.
2. Stage 1: ½ × 1.5 × 8.0² = 48 m. Then stage 2: 12 × 20.0 = 240 m. Stage 3: ½ × 12 × 5.0 = 30 m. Total = 318 m.
3. a = (0 − 12)/5.0 = −2.4 m/s². Total time = 8.0 + 20.0 + 5.0 = 33 s, so average velocity = 318/33 = 9.6 m/s.

Problem 15 Doubling the speed

1. Δx = 15²/(2 × 5.0) = 22.5 m; t = 15/5.0 = 3.0 s.
2. Δx = 30²/10 = 90 m; t = 30/5.0 = 6.0 s.
3. With v = 0, Δx is proportional to v₀², so doubling the speed gives four times the distance, while t = v₀/a is proportional to v₀, so the time only doubles.

Problem 16 Acceleration-time graph of a sled

1. Area gives Δv: 4.0 × 5 = 20 m/s at 5 s; no change until 8 s, so 20 m/s; then −2.0 × 6 = −12, so 8.0 m/s at 14 s.
2. 0–5 s: ½ × 20 × 5 = 50 m; 5–8 s: 20 × 3 = 60 m; 8–14 s: ½ × (20 + 8.0) × 6 = 84 m. Total = 194 m.
3. A straight line rising from 0 to 20 m/s over 5 s, a flat line to 8 s, then a straight line falling to 8.0 m/s at 14 s.

Problem 17 Sandbag from a balloon

1. v₀ = +4.0 m/s (it keeps the balloon’s velocity) and a = −9.81 m/s².
2. −60.0 = 4.0t − 4.905t², so 4.905t² − 4.0t − 60.0 = 0. t = [4.0 + √(16 + 4 × 4.905 × 60.0)]/9.81 = (4.0 + 34.54)/9.81 = 3.93 s.
3. v = 4.0 − 9.81 × 3.93 = −34.5 m/s (34.5 m/s downward). It rises 4.0²/(2 × 9.81) = 0.82 m first, so the greatest height is 60.8 m.

Problem 18 Design: acceleration of a cart

1. Release the cart from rest at the same start line each time. Moreover, measure the distance Δx from the start line to marks at several positions, and the time t for each with several repeats. Also, keep the ramp angle and the release point the same.
2. Then plot Δx (vertical) against t² (horizontal). Since Δx = ½at² with v₀ = 0, the graph is a straight line through the origin and a = 2 × slope.
3. Time each distance at least three times and average, or use longer distances so the reaction time of about 0.2 s is a smaller fraction of t.

Problem 19 Argument: acceleration at the top

1. Each 0.10 s step changes v by −0.98 m/s, so a = −0.98/0.10 = −9.8 m/s² for every pair.
2. The claim is wrong. The velocity passes through zero near t = 1.22 s, but it is changing at the same rate of −9.8 m/s² before, at and after that instant: it goes from +0.23 to −0.75 m/s in 0.10 s. Acceleration is the rate of change of velocity, not the velocity, so a ball momentarily at rest at the top still has a = 9.81 m/s² downward.

Problem 20 Sprinter split times

1. Section times: 1.85, 1.02, 0.93, 0.85, 0.82, 0.82 s, so v = 5.4, 9.8, 10.8, 11.8, 12.2, 12.2 m/s.
2. The last two sections have the same average velocity, so she reaches top speed about 12.2 m/s at roughly 40–50 m.
3. v = 60/6.29 = 9.5 m/s.
4. With a constant acceleration from rest, the first-section average would be 5.4 m/s, meaning a = 2 × 10/1.85² = 5.8 m/s² and v = 10.8 m/s at 10 m. A constant a would then give the second section an average above 10.8 m/s, but only 9.8 m/s is measured, so the acceleration decreases.