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 $30 \,s$.
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 $30 \,s$.
Given $d=49 m , T =20 s , t=30 s$
At the end of $30 s$, rounds completed are
$n=\frac{30}{20}=1.5$
Therefore, distance travelled
$S=2 \pi r=2 \times \frac{22}{7} \times \frac{49}{2} \times 1.5=231 m$
Displacement is equal to the diameter $49 m$ as after the completion of $1.5$ rounds the athlete will be diametrically opposite to his point of start.
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