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A circular cycle track has a circumference of $314\, m$ with $A B$ as one of its diameter. $A$ cyclist travels from $A$ to $B$ along the circular path with a velocity of constant magnitude $15.7\, m s ^{-1}$. Find the
$(a)$ distance moved by the cyclist.
$(b)$ displacement of the cyclist, if $A B$ represents north$-$south direction.
$(c)$ the average velocity of the cyclist.
Solution
$(a)$ Distance travelled by the cyclist $=$ length of the path $AB$ $=$ half the circumference of the circle
$S=\frac{314}{2}=157 m$
$(b)$ The displacement of the cyclist is equal to the diameter $AB$ of the circle.
Now, circumference $=2 \pi r=314,$ therefore,
$r=\frac{\text { circumference }}{2 \pi}=\frac{314}{2 \times 3.14}=50 m$
Therefore, displacement of the cyclist $=2 \times$ $50=100 \,m$ towards south.
$(c)$ Since the magnitude of velocity is constant, therefore, the average velocity is equal to the velocity with which the cyclist moves. Therefore, average velocity $=15.7 m s ^{-1}$
