Practice questions · Motion graphs

Gradients and areas.

Six original Cambridge-style questions on reading both graph types and on finding distance as the area under a speed-time line, split into triangles and rectangles.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The two graphs, side by side

Gradient is rate; area is distance.

Distance-time: gradient = speed; a flat line means stationary.
Speed-time: gradient = acceleration; area under the line = distance.
Changing speed: split the area into triangles and rectangles, then add. Never use peak speed × total time.

Analysis

[2 marks]

On a distance-time graph, state what is represented by:

(a)the gradient of the line; (b)a horizontal (flat) line.

(a) The gradient represents the speed. ✓
(b) A flat line means the object is stationary (not moving). ✓

Analysis

[2 marks]

On a speed-time graph, state what is represented by:

(a)the gradient of the line; (b)the area under the line.

(a) The gradient represents the acceleration. ✓
(b) The area under the line represents the distance travelled. ✓

Calculation

[3 marks]

On a speed-time graph, a car accelerates uniformly from rest to 15 m/s in 6.0 s, then travels at a steady 15 m/s for 10 s. Calculate the total distance travelled.

Triangle. ½ × 6.0 × 15 = 45 m

Rectangle. 15 × 10 = 150 m

total = 195 m

Calculation

[4 marks]

A cyclist accelerates from rest to 8.0 m/s in 4.0 s, holds 8.0 m/s for 6.0 s, then slows uniformly to rest in 2.0 s. Calculate the total distance travelled.

Speeding up. ½ × 4.0 × 8.0 = 16 m

Steady. 8.0 × 6.0 = 48 m

Slowing. ½ × 2.0 × 8.0 = 8 m

total = 72 m

Analysis

[2 marks]

To find the distance from a speed-time graph where the speed changes, a student multiplies the final speed by the total time. Explain why this is wrong and state the correct method.

That treats the whole graph as one rectangle, but the object was not at the final speed for the whole time, so it over-counts the distance. ✓
Correct method: find the area under the line, splitting it into triangles and rectangles and adding them. ✓

Calculation

[3 marks]

On a distance-time graph, an object's line rises steadily from 0 to 100 m over the first 8.0 s, then stays flat at 100 m for the next 5.0 s.

(a)Calculate the speed during the first 8.0 s. (b)State what the object is doing during the flat section.

(a) speed = gradient = 100 m ÷ 8.0 s

= 12.5 m/s

(b) It is stationary (staying at the same distance). ✓