The velocity-time graph from 0 s to 120 s for a cyclist is shown in the given figure.
Shade the areas (in different colours) representing the displacement of the cyclist:
(A) while cyclist is moving with constant velocity.
(B) when the velocity of cyclist is decreasing. Also, calculate the displacement and average acceleration in the 120 s time interval.

(A) In the given velocity-time graph, shade the rectangular area under the graph from 20 s to 100 s in one colour. This area represents the displacement while the cyclist is moving with constant velocity.

(B) Shade the trapezium-shaped area under the graph from 100 s to 120 s in another colour. This area represents the displacement while the cyclist is moving with decreasing velocity.

The displacement of the cyclist is equal to the total area under the velocity-time graph.

$$\text{Displacement} = \frac{1}{2} \times 20 \times 3 + 80 \times 3 + \frac{1}{2}(3+2) \times 20$$

$$= 30 + 240 + 50$$

$$= 320 \text{ m}$$

Now,

$$u = 0 \text{ ms}^{-1}, v = 2 \text{ ms}^{-1}, t = 120 \text{ s}$$

$$\text{Using, } a = \frac{v-u}{t}$$

$$a = \frac{2}{120}$$

$$a = 0.017 \text{ ms}^{-2}$$

The displacement of the cyclist is 320 m and the average acceleration is 0.017 ms$^{-2}$.