Answers · Speed and motion graphs

Speed and motion graphs — final check key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. MK1 · Final check

Model answer. (a) 0–2 s at rest at 0.5 m; 2–6 s steady speed away from the sensor; 6–9 s at rest at 1.7 m; 9–11 s steady speed back towards the sensor; 11–12 s at rest at 0.9 m. (b) Speed = 1.2 m ÷ 4 s = 0.30 m/s. (c) The distance falls by 0.8 m in 2 s: velocity = −0.40 m/s and speed = 0.40 m/s. Velocity is speed in a given direction: the minus sign says the car is moving towards the sensor; speed is only the size. (d) Distance travelled = 1.2 m + 0.8 m = 2.0 m; average speed = 2.0 m ÷ 12 s = 0.17 m/s. (e) Continue the graph: flat at 0.9 m to 12 s; a straight line rising to 0.9 m + 0.20 m/s × 3 s = 1.5 m at 15 s (a steady speed away from the sensor); flat at 1.5 m from 15 s to 18 s (at rest).

A full answer includes

(a) At rest; steady speed away from the sensor; at rest; steady speed back towards the sensor; at rest.
(b) (1.7 − 0.5) m ÷ 4 s = 0.30 m/s.
(c) Velocity −0.40 m/s, speed 0.40 m/s (0.8 m in 2 s, towards the sensor): velocity is speed in a given direction, so its sign shows the car is coming back; speed has no direction.
(d) Distance travelled 1.2 m + 0.8 m = 2.0 m; average speed 2.0 m ÷ 12 s = 0.17 m/s.
(e) The line stays flat at 0.9 m until 12 s, rises in a straight line to 1.5 m at 15 s (0.20 m/s × 3 s = 0.6 m further), then is flat at 1.5 m until 18 s; the time axis reaches 18 s.

Also accept

(b) 0.30 m/s; (c) −0.40 m/s, speed 0.40 m/s; (d) 2.0 m, 0.17 m/s; (e) straight rise to 1.5 m at 15 s, then flat

Watch for

“(d) The car ends 0.4 m further out, so the distance travelled is 0.4 m.” — That is the change in distance from the sensor. The car went 1.2 m out and 0.8 m back: 2.0 m travelled.
“(c) The velocity is 0.40 m/s.” — That is the speed. The car moves towards the sensor, so its velocity is negative: −0.40 m/s.

2. MK2 · Final check

Model answer. (a) 0–2 s: speeds up evenly from rest to 3 m/s. 2–7 s: constant speed of 3 m/s. 7–10 s: slows down evenly to rest. 10–12 s: at rest. (b) Distance = ½ × 2 s × 3 m/s + 5 s × 3 m/s + ½ × 3 s × 3 m/s = 3 m + 15 m + 4.5 m = 22.5 m. (c) Average speed = 22.5 m ÷ 12 s = 1.9 m/s. (d) Distance–time shape: 0–2 s a curve that gets steeper (speeding up); 2–7 s a straight rising line (constant speed); 7–10 s a curve that gets flatter (slowing down); 10–12 s flat at 22.5 m (at rest).

A full answer includes

(a) Speeds up evenly from rest to 3 m/s; constant speed 3 m/s; slows down evenly to rest; at rest.
(b) ½ × 2 × 3 + 5 × 3 + ½ × 3 × 3 = 3 + 15 + 4.5 = 22.5 m (23 m).
(c) 22.5 m ÷ 12 s = 1.9 m/s.
(d) A curve getting steeper (0–2 s), a straight rising line (2–7 s), a curve getting flatter (7–10 s), then flat at 22.5 m (10–12 s); labelled axes.

Also accept

(b) 22.5 m; (c) 1.9 m/s; (d) steeper curve, straight line, flatter curve, flat

Watch for

“(b) 3 m/s × 10 s = 30 m.” — The lift only moved at 3 m/s for 5 s. Use triangles for speeding up and slowing down.
“(c) Divide by 10 s because the lift stops at 10 s.” — Average speed over the 12 s uses the whole 12 s, including the time at rest.