A cyclist starts form centre ` O` of a circular park of radius ` 1 km` and moves along the path ` OPRQO`. 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, accelertion is called centripetal acceleration and is always directed toward the centre.
Hence, acceleration at R is a =`(v^(2))/(r)`
`rArr " " a=((10)^(2))/(1km)=(100)/(10^(3))=0.1m//s^(2)`along RO.