A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the (i) Net displacement (ii) Average velocity and (iii) The average speed of the cyclist.
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the (i) Net displacement (ii) Average velocity and (iii) The average speed of the cyclist.

Answer –

(i) Displacement refers to the distance between the body’s original and ultimate positions. In 20 minutes, the cyclist returns to the point where he began. As a result, there is no displacement.

(ii) Average velocity = (net displacement) / (time taken)

(iii) Average speed is given by

Sped avg. = distance travelled / time taken

In this case, from the diagram above, we have upon putting values

= OP + Distance PQ + QO/ 10 minutes

= {1 km + (1/4) x 2 x (22/7) x 1km + 1m}/ (10/60) h

= 6 (2 + 22/14) = 6 (50/14)

Average speed = 21.43 km/h