$A$ 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 Figure. If the round trip takes $10 \;min$,what is the
$(a)$ net displacement,
$(b)$ average velocity,and
$(c)$ average speed of the cyclist?

(N/A) Displacement is defined as the shortest distance between the initial and final positions. Since the cyclist starts at $O$ and returns to $O$ after the round trip,the initial and final positions are the same. Therefore,the net displacement is $0 \; km$.
$(b)$ Average velocity is defined as the ratio of net displacement to total time taken:
$\text{Average velocity} = \frac{\text{Net displacement}}{\text{Total time}}$
Since the net displacement is $0$,the average velocity is $0 \; km/h$.
$(c)$ Average speed is defined as the ratio of total path length to total time taken:
$\text{Average speed} = \frac{\text{Total path length}}{\text{Total time}}$
The total path length is the sum of the distances $OP$,$PQ$ (arc length),and $QO$:
$OP = 1 \; km$
$PQ = \frac{1}{4} \times (2 \pi r) = \frac{1}{4} \times 2 \times \pi \times 1 = \frac{\pi}{2} \approx 1.57 \; km$
$QO = 1 \; km$
$\text{Total path length} = 1 + 1.57 + 1 = 3.57 \; km$
$\text{Total time} = 10 \; min = \frac{10}{60} \; h = \frac{1}{6} \; h$
$\text{Average speed} = \frac{3.57}{1/6} = 3.57 \times 6 = 21.42 \; km/h$