Reading a whole journey off one graph

Now let us read a three-stage journey off one graph.

The problem

An object starting from rest undergoes a uniform acceleration for 6 seconds and reaches a velocity of 15 m s⁻¹. It then moves at that uniform velocity for another 6 seconds, and finally undergoes a uniform deceleration, coming to rest in 3 seconds.

t (s)	0	6	12	15
v (m s⁻¹)	0	15	15	0

The rise from O to A, the flat section from A to B, and the fall from B to C: all three stages are visible at once.

(i) Acceleration in the first 6 s

This is the gradient of the line OA.

15 m s⁻¹ ÷ 6 s = 2.5 m s⁻²

(ii) Displacement in the first 6 s

This is the area of the triangle under OA.

(15 × 6) ÷ 2 = 45 m

(iii) Distance at uniform velocity

This is the area of the rectangle under AB.

15 m s⁻¹ × 6 s = 90 m

(iv) Deceleration in the last 3 s

acceleration = (0 − 15) m s⁻¹ ÷ 3 s = −5 m s⁻²

So the deceleration is 5 m s⁻².

(v) Distance in the last 3 s

((15 + 0) ÷ 2) × 3 s = 22.5 m

(vi) Total distance

The three answers can simply be added: 45 + 90 + 22.5 = 157.5 m.

But there is a shorter way. The whole shape is a trapezium OABC, so its area can be found in one step:

area of a trapezium = (sum of the two parallel sides ÷ 2) × height

parallel sides = 15 s (the bottom) and 6 s (the top)

((15 + 6) ÷ 2) × 15 = (21 ÷ 2) × 15 = 157.5 m

What the graph buys you

Instead of three separate calculations, the whole answer comes from the area of a single geometric shape. The more complicated the motion, the more the graph is worth.