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

Explanation

As shown in the adjacent figure. The cyclist covers the path OPRQO. As we know whenever an object performing circular motion, acceleration is called centripetal acceleration and is always directed towaeds the centre.
Hence, acceleration at R is `a=(v^2)/(r)`
`impliesa=((10)^2)/(1km)=(100)/(10^3)=0.1(m)/(s^2)` along RO.