The story of a journey — answer key

1. MQ1 · Practice

Model answer. Speed = distance ÷ time = 1.2 m ÷ 6.0 s = 0.20 m/s.

Checked automatically

• Speed of the trolley: 0.20 m/s (within ±2%)
  • Common wrong answer 5 : You divided the time by the distance. Speed is distance ÷ time.
  • Common wrong answer 7.2 : You multiplied. Speed is distance ÷ time.

A full answer includes

• I used speed = distance ÷ time: 1.2 m ÷ 6.0 s = 0.20 m/s, with the unit m/s.

Also accept

• 0.20 m/s (0.2 m/s)

Watch for

• “Speed = time ÷ distance = 5 m/s.” — That is the number of seconds for each metre. Speed is metres per second: divide the distance by the time.
• “Speed = distance × time = 7.2 m/s.” — Multiplying gives a bigger number for a slower trolley that takes longer. Divide the distance by the time.

2. MQ2 · Practice

Model answer. b. Both move at 0.3 m/s, so their speeds are equal, but they move in opposite directions, so their velocities are different: taking away from the sensor as positive, A has a velocity of +0.3 m/s and B has −0.3 m/s. Velocity is speed in a given direction; speed is only the size and is never negative.

A full answer includes

• I chose b: same speed, different velocities.
• I explained that velocity is speed in a given direction; speed has no direction, so it is never negative.

Also accept

• b; velocity is speed in a given direction.

Watch for

• “They have the same velocity because they move at the same speed.” — Velocity includes the direction. Opposite directions mean different velocities, even at the same speed.
• “B has a speed of −0.3 m/s.” — Speed has no direction and is never negative. The minus sign belongs to B's velocity.

3. MQ3 · Practice

Model answer. Total distance = 600 m + 900 m = 1500 m. Total time = 2.0 + 1.0 + 2.0 = 5.0 minutes = 300 s, including the wait, because the clock keeps running. Average speed = 1500 m ÷ 300 s = 5.0 m/s.

Checked automatically

• Average speed: 5.0 m/s (within ±2%)
  • Common wrong answer 6.3 : Include the minute spent at the lights in the total time, or don't average the two speeds: use total distance ÷ total time.
  • Common wrong answer 300 : 1500 m in 5.0 minutes is 300 m per minute. Change the time to seconds to get m/s.

A full answer includes

• I added the distances: 600 m + 900 m = 1500 m.
• I added all the times, including the wait: 5.0 minutes = 300 s.
• I divided: 1500 m ÷ 300 s = 5.0 m/s.

Also accept

• 5.0 m/s (5 m/s)

Watch for

• “Leave out the minute at the lights: 1500 m ÷ 240 s = 6.3 m/s.” — The journey took 5 minutes from start to finish. Average speed uses the total time, stops included.
• “Average the two riding speeds: (5 m/s + 7.5 m/s) ÷ 2 = 6.3 m/s.” — That ignores the stop and the time spent at each speed. Use total distance ÷ total time.

4. MQ4 · Practice

Options as numbered on the pupil sheet: 1 — Speeding up; 2 — Moving at a steady speed away from the start; 3 — At rest; 4 — Moving at a steady speed back towards the start.

Model answer. A: speeding up, because the curve gets steeper (more distance each second). B: steady speed away from the start, because the line is straight and rising. C: at rest, because the line is flat (the distance doesn't change). D: steady speed back towards the start, because the line is straight and falling (the distance from the start gets smaller).

A full answer includes

• I matched A: speeding up; B: steady speed away from the start; C: at rest; D: steady speed back towards the start.

Also accept

• A speeding up; B steady away; C at rest; D steady back.

Watch for

• “D is slowing down because the line goes down.” — A falling line on a distance–time graph means the distance from the start is getting smaller: she is coming back. Slowing down would make the line flatter.
• “C is a steady speed because the line is straight.” — A flat line means the distance isn't changing at all: she is at rest.

5. MQ5 · Practice

Model answer. Axes: time / s across, distance from the owner / m up. From 0 to 5 s a flat line at 2 m (at rest). From 5 s to 10 s a straight line rising from 2 m to 22 m, because the dog covers 4 m/s × 5 s = 20 m. From 10 s to 15 s a flat line at 22 m (at rest again).

A full answer includes

• My axes are labelled time / s and distance from the owner / m.
• My graph is flat at 2 m from 0 to 5 s, then a straight line rising to 22 m at 10 s, then flat at 22 m until 15 s.
• I worked out the end distance: 2 m + 4 m/s × 5 s = 22 m.

Also accept

• flat at 2 m to 5 s; straight rise to 22 m at 10 s; flat at 22 m to 15 s; labelled axes.

Watch for

• “Start the graph at 0 m.” — The dog starts 2 m from its owner, so the line starts at 2 m on the distance axis.
• “Draw the running part as a curve.” — At a steady speed the distance grows by the same amount every second, so the line is straight.

6. MQ16 · Practice

Model answer. Change in distance = 25 m − 5 m = 20 m; change in time = 6 s − 2 s = 4 s. Gradient = 20 m ÷ 4 s = 5.0 m/s. The gradient of a distance–time graph is the speed, so the skateboarder moves at a steady 5.0 m/s.

Checked automatically

• Gradient (speed): 5.0 m/s (within ±2%)
  • Common wrong answer 4.2 : You divided the final distance by the final time. Use the changes between the two points.
  • Common wrong answer 0.20 : You divided the time by the distance. Gradient = change in distance ÷ change in time.

A full answer includes

• I found the changes: 25 m − 5 m = 20 m, and 6 s − 2 s = 4 s.
• I divided: 20 m ÷ 4 s = 5.0 m/s, and said the gradient is the skateboarder's speed.

Also accept

• 5.0 m/s; it is the speed

Watch for

• “25 m ÷ 6 s = 4.2 m/s.” — The line doesn't start at 0 m and 0 s. Use the change between the two points: 20 m in 4 s.
• “4 ÷ 20 = 0.20 m/s.” — That is time ÷ distance. Gradient = change in distance ÷ change in time.

7. MQ6 · Practice

Model answer. (a) Gradient = 60 m ÷ 40 s = 1.5 m/s. (b) On the way back the distance falls by 60 m between 70 s and 100 s: 60 m ÷ 30 s = 2.0 m/s. Her velocity is −2.0 m/s (towards the door), and her speed is 2.0 m/s.

Checked automatically

• (a) Speed during the first 40 s: 1.5 m/s (within ±2%)
  • Common wrong answer 0.67 : You divided the time by the distance. Gradient = change in distance ÷ change in time.
• (b) Speed on the way back: 2.0 m/s (within ±2%)
  • Common wrong answer 0.60 : The way back lasts from 70 s to 100 s. Divide by that change in time, 30 s.
  • Common wrong answer -2.0 : That is her velocity (towards the door). Speed is just the size, without a sign.
• (c) Velocity on the way back (away from the door positive): -2.0 m/s (within ±2%)
  • Common wrong answer 2.0 : 2.0 m/s is her speed. She walks towards the door, the negative direction, so her velocity is −2.0 m/s.

A full answer includes

• (a) I read 60 m in 40 s and found 60 ÷ 40 = 1.5 m/s.
• (b) I read 60 m in 30 s (from 70 s to 100 s) and found 2.0 m/s; I gave the speed without a minus sign (her velocity is −2.0 m/s).
• (c) Her velocity on the way back is −2.0 m/s: the same size as her speed, with a minus sign because she walks towards the door. Velocity is speed in a given direction.

Also accept

• (a) 1.5 m/s; (b) 2.0 m/s; (c) −2.0 m/s

Watch for

• “Use 100 s for the way back: 60 ÷ 100 = 0.60 m/s.” — The way back takes from 70 s to 100 s: 30 s. Use the change in time for that section only.
• “Her speed on the way back is −2.0 m/s.” — −2.0 m/s is her velocity (towards the door). Speed has no direction: 2.0 m/s.

8. MQ7 · Practice

Model answer. Total distance = 1800 m + 2700 m = 4500 m. Total time = 3.0 + 2.0 + 5.0 = 10.0 minutes = 600 s. Average speed = 4500 m ÷ 600 s = 7.5 m/s.

Checked automatically

• Average speed of the bus: 7.5 m/s (within ±2%)
  • Common wrong answer 9.4 : Include the 2.0 minutes at the stop in the total time.
  • Common wrong answer 9.5 : Don't average the two speeds: the bus also stood still. Use total distance ÷ total time.
  • Common wrong answer 0.45 : That is 4.5 km in 10 minutes, in km per minute. Convert to metres and seconds.

A full answer includes

• I found the total distance: 1800 m + 2700 m = 4500 m.
• I found the total time including the stop: 10.0 minutes = 600 s.
• I divided: 4500 m ÷ 600 s = 7.5 m/s.

Also accept

• 7.5 m/s

Watch for

• “Leave out the stop: 4500 m ÷ 480 s = 9.4 m/s.” — The stop is part of the journey time. Average speed = total distance ÷ total time, stops included.
• “Average the two speeds: (10 m/s + 9 m/s) ÷ 2 = 9.5 m/s.” — That ignores the stop and the different times at each speed. Use the totals.

9. MQ8 · Practice

Model answer. a, c and d. In each second the car moves 0.10, 0.30, 0.50, 0.70 and 0.90 m: the gaps grow, so it is speeding up (a), and its distance–time graph is a curve that gets steeper (c). Between 4 s and 5 s it moves 0.90 m in 1 s, so its average speed in that second is 0.90 m/s (d). It is not moving at a steady speed (b), and its average speed over the whole 5 s is 2.50 m ÷ 5 s = 0.50 m/s, not 2.50 m/s (e).

A full answer includes

• I chose a, c and d.
• I explained that the car moves further in each second (0.10, 0.30, 0.50, 0.70, 0.90 m), so it is speeding up; I did not choose b or e.

Also accept

• a, c, d

Watch for

• “The distances go up evenly, so the speed is steady.” — Look at the differences, not the distances: 0.10, 0.30, 0.50 … m each second. Growing gaps mean speeding up.
• “The average speed is 2.50 m/s because it reached 2.50 m.” — 2.50 m is a distance. Divide by the total time, 5 s: 0.50 m/s.

10. MQ9 · Practice

Model answer. (a) Time / s across (0 to 10 s), distance from the sensor / m up (0 to 1.2 m); six crosses joined part by part with straight lines. (b) 0–2 s: flat, at rest at 0.30 m. 2–6 s: straight and rising, a steady speed away from the sensor. 6–8 s: flat, at rest at 1.10 m. 8–10 s: falling, coming back towards the sensor (0.60 m in 2 s, 0.30 m/s). (c) Speed = (1.10 − 0.30) m ÷ (6 − 2) s = 0.80 m ÷ 4 s = 0.20 m/s.

A full answer includes

• (a) My axes are labelled time / s and distance from the sensor / m, the points are small crosses, and I joined them with straight lines part by part.
• (b) At rest 0–2 s; steady speed away 2–6 s; at rest 6–8 s; moving back towards the sensor 8–10 s.
• (c) Gradient = (1.10 − 0.30) m ÷ (6 − 2) s = 0.20 m/s.

Also accept

• (b) rest, steady away, rest, back; (c) 0.20 m/s

Watch for

• “Join all the points with one smooth curve.” — This trolley's motion has separate parts with corners between them. Join the points part by part.
• “Speed between 2 s and 6 s = 1.10 m ÷ 6 s = 0.18 m/s.” — The trolley started that section at 0.30 m and at 2 s. Use the change in distance (0.80 m) over the change in time (4 s) for that section only.

11. MT1 · Apply

Model answer. (a) The vertical axis is her distance from the starting wall. On the second length she swims towards the wall, so that distance gets smaller and the line falls, even though she is still swimming. (b) At the end she is 0 m from the wall, but she has travelled 100 m. (c) Average speed = total distance travelled ÷ total time = 100 m ÷ 80 s = 1.25 m/s (1.3 m/s to 2 significant figures). (d) The falling part drops 50 m in 40 s: gradient = (0 − 50) m ÷ 40 s = −1.25 m/s, so her velocity is −1.25 m/s (towards the wall) and her speed is 1.25 m/s.

A full answer includes

• (a) I explained that the graph shows distance from the wall, which gets smaller as she swims back towards it.
• (b) I said 0 m from the wall, but 100 m travelled.
• (c) I calculated 100 m ÷ 80 s = 1.25 m/s (1.3 m/s).
• (d) I found the gradient (0 − 50) m ÷ 40 s = −1.25 m/s: her velocity is −1.25 m/s (1.25 m/s towards the wall).

Also accept

• (a) distance from the wall decreases; (b) 0 m, 100 m; (c) 1.25 m/s; (d) −1.25 m/s

Watch for

• “Her average speed is zero because she ends where she started.” — She swam 100 m. Average speed uses the distance travelled, not where she ends up.
• “The line falls because she slows down.” — Slowing down would make the line less steep. A falling line means the distance from the wall is getting smaller.