Question

The velocity-time graph from 0 s to 120 s for a cyclist is shown in the Fig. Shade the areas (in different colours) representing the displacement of the cyclist

1. while cyclist is moving with constant velocity.
2. when the velocity of cyclist is decreasing.

Also, calculate the displacement and average acceleration in the 120 s time interval.

Answer 1

(i) Constant Velocity: Shade the rectangular area between t =20 s and t = 100 s

(ii) Decreasing Velocity: Shade the trapezoidal area between t = 100 s and t = 120 s (where the line slopes downward).

Displacement Calculation (Total Area under the Graph):

Area 1 (Triangle, 0 – 20s):

`1/2 xx 20 xx 3`

= 30 m

Area 2 (Rectangle, 20 – 100s):

length × width

= 80 × 3

= 240 m

Area 3 (Trapezium, 100 – 120s):

`1/2 xx (3 + 2) xx 20`

= 50 m

Total Displacement: 30 + 240 + 50 = 320 m

Average Acceleration = `["Final velocity − Initial Velocity"/"Total Time"]`

= `(2 - 0)/120`

= `1/60` m/s²