Distance–Time and Velocity–Time Graphs — Answers
Answer sheet with model answers and success criteria for the Year 11 forces & motion worksheet.
Success TutoringWorksheetsAnswer sheetYear 11 · Science · Forces & MotionDistance–Time and Velocity–Time Graphs — Answers
Part A: Reading graph shapes
On a distance–time graph, the gradient (steepness) is the speed. A straight sloping line means constant speed; a steeper line means a faster speed; a flat line means the object is stopped.
On a velocity–time graph, the gradient is the acceleration, and the area under the line is the distance travelled. A line sloping up means speeding up; sloping down means slowing down; flat means constant velocity.
Gradient = rise ÷ run (change in the y-value ÷ change in time).
- Amoving at a constant speed
- Bstationary
- Caccelerating
- Dslowing down
Answer: stationary
The distance is not changing, so the object is not moving.
- Astationary
- Bmoving at a constant velocity
- Caccelerating
- Ddecelerating
Answer: moving at a constant velocity
The velocity is not changing, so the acceleration is zero, but the object is still moving.
Answer: The object is speeding up (accelerating), because the gradient, which is the speed, is increasing.
Part B: Distance–time graph
A cyclist's distance–time graph has three straight sections: from 0 s to 10 s the line rises from 0 m to 50 m; from 10 s to 20 s it is flat at 50 m; from 20 s to 30 s it rises from 50 m to 150 m.
Answer: 5 m/s
- speed = gradient = 50 ÷ 10 = 5 m/s
Answer: The cyclist is stationary (stopped) at 50 m from the start: the distance does not change.
Answer: 10 m/s
- Distance = 150 − 50 = 100 m in 10 s
- 100 ÷ 10 = 10 m/s
Answer: 5 m/s
- average speed = total distance ÷ total time
- 150 ÷ 30 = 5 m/s
Part C: Velocity–time graph
A car's velocity–time graph: it starts from rest and its velocity rises steadily to 20 m/s in 8 s; it then travels at a constant 20 m/s for 12 s; finally it slows steadily to a stop in 5 s.
Answer: 2.5 m/s²
- a = change in velocity ÷ time
- 20 ÷ 8 = 2.5 m/s²
Answer: −4 m/s² (a deceleration of 4 m/s²)
- a = (0 − 20) ÷ 5 = −4 m/s²
Answer: 370 m
- Stage 1 (triangle): ½ × 8 × 20 = 80 m
- Stage 2 (rectangle): 12 × 20 = 240 m
- Stage 3 (triangle): ½ × 5 × 20 = 50 m
- Total = 80 + 240 + 50 = 370 m
Answer: 14.8 m/s
- Total time = 8 + 12 + 5 = 25 s
- 370 ÷ 25 = 14.8 m/s
Answer: False
The maximum speed is 20 m/s, but the average speed is only 14.8 m/s, because the car was slower than 20 m/s while speeding up and slowing down.
Part D: Stopping distances
Answer: 14 m
- distance = speed × time
- 20 × 0.7 = 14 m
Answer: 4 s; 40 m
- time = 20 ÷ 5 = 4 s
- braking distance = ½ × 4 × 20 = 40 m
Answer: 54 m; 70 m
- Alert: 14 + 40 = 54 m
- Tired: thinking distance = 20 × 1.5 = 30 m; 30 + 40 = 70 m
Model answer: (a) Using a phone distracts the driver, so the reaction time is longer. The car travels further before braking, so the thinking distance increases; the braking distance is unchanged. (b) A wet road reduces the friction between the tyres and the road, so the brakes cannot decelerate the car as much, and the braking distance increases; the thinking distance is unchanged. (c) At a higher speed, the car travels further during the reaction time, so the thinking distance increases, and it takes longer to stop, so the braking distance also increases (braking distance increases with the square of the speed). Both distances increase.
Success criteria:
- Phone: longer reaction time, thinking distance increases
- Wet road: less friction, braking distance increases
- Higher speed increases both thinking and braking distance
- Clear explanations linked to reaction time or friction
Marking tips
- Give credit for correct method even when the final answer slips — the working shows where the thinking went right.
- For open-ended tasks, use the success criteria as a checklist rather than looking for one "right" answer.
- Celebrate what went well first, then pick one thing to work on next.