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3.12c · Rates of Change · Sub-skill
Use distance–time and speed–time graphs: gradient for speed and acceleration, area for distance.
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Build it up, step by step
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On a distance–time graph the gradient is the speed. A straight line means constant speed, a flat line means stopped, and a steeper line means faster. Units: metres per second (m/s), kilometres per hour (km/h).
On a speed–time graph the gradient is the acceleration (units m/s²). A positive gradient means speeding up, a flat line means constant speed, and a negative gradient means slowing down (deceleration).
The area under a speed–time graph is the distance travelled. Split the area into rectangles, triangles and trapeziums: a rectangle is speed × time; a triangle is ½ × base × height.
When a graph is curved the rate is not constant. The speed (or acceleration) at an instant is the gradient of the tangent at that point. The area under a curve can be estimated by counting squares or using trapeziums.
A vehicle speeds up from 0 to 20 m/s in 4 s, travels at 20 m/s for 4 s, then slows to a stop in 4 s. Find the acceleration at the start and the total distance.
Acceleration $= \dfrac{20 - 0}{4} = $ 5 m/s².
Distance = area under graph = $\frac{1}{2} \times 4 \times 20 + 4 \times 20 + \frac{1}{2} \times 4 \times 20 = 40 + 80 + 40 = $ 160 m.
Test yourself
A car accelerates uniformly from rest to 12 m/s in 6 seconds, then travels at 12 m/s for 10 seconds.
(a) Find the acceleration in the first 6 seconds.
(b) Find the distance travelled in the first 6 seconds.
(c) Find the total distance travelled in the 16 seconds. [5 marks]
(a) $\dfrac{12 - 0}{6} = $ 2 m/s². (1 mark)
(b) Area of triangle = $\frac{1}{2} \times 6 \times 12 = $ 36 m. (2 marks)
(c) Second stage: $12 \times 10 = 120$ m. Total = 36 + 120 = 156 m. (2 marks)
Practice
Five short questions. Work through them, then reveal the mark scheme to check.