An athlete completes one round of a circular track of diameter $49 \,m$ in $20 \,s$. Calculate the distance covered and displacement at the end of $30 \,s$.

(DISTANCE = 231 M, DISPLACEMENT = 49 M) Given: Diameter $d = 49 \,m$,Time for one round $T = 20 \,s$,Total time $t = 30 \,s$.
First,calculate the number of rounds completed in $30 \,s$:
$n = \frac{t}{T} = \frac{30}{20} = 1.5 \, \text{rounds}$.
Distance covered is the total path length:
$S = n \times (2 \pi r) = n \times (\pi d) = 1.5 \times \frac{22}{7} \times 49 = 1.5 \times 22 \times 7 = 231 \,m$.
Displacement is the shortest distance between the initial and final positions:
After $1.5$ rounds,the athlete is at the point diametrically opposite to the starting point.
Therefore,the displacement is equal to the diameter of the circular track,which is $49 \,m$.