$A$ cyclist starts from the centre $O$ of a circular park of radius $1\, km$ and moves along the path $OPRQO$ as shown in the figure. If he maintains a constant speed of $10\, m/s$,what is his acceleration at point $R$ in magnitude and direction?

(N/A) The cyclist is moving along a circular path with a constant speed. Therefore,the motion is uniform circular motion.
In uniform circular motion,the acceleration is purely centripetal,directed towards the centre of the circle.
The magnitude of centripetal acceleration is given by $a = \frac{v^2}{r}$.
Given: speed $v = 10\, m/s$ and radius $r = 1\, km = 1000\, m$.
Substituting the values:
$a = \frac{(10\, m/s)^2}{1000\, m} = \frac{100}{1000}\, m/s^2 = 0.1\, m/s^2$.
The direction of this acceleration is always towards the centre of the circular path,which is point $O$. Thus,the acceleration is $0.1\, m/s^2$ directed along $RO$.