A cyclist starts form centre ` O` of a circular park of radius ` 1 km` and moves along the path ` OPRQO` as shown Fig. 2 (EP).15. If he maintains constant speed of ` 10 ms^(-1)`, what is his acceleration at point (R )in magnitude and direction ?
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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.