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प्रश्न
A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown figure. If he maintains constant speed of 10 ms–1, what is his acceleration at point R in magnitude and direction?
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उत्तर
According to the problem, the path of the cyclist is OPRQO.
The cyclist is in uniform circular motion and it is given that linear velocity = 10 m/s, R = 1 km = 1000 m. As we know whenever an object is performing a circular motion, acceleration is called centripetal acceleration and is always directed towards the centre. So cyclist experiences a centripetal force (acceleration) at point R towards the centre.
The centripetal acceleration at R is given by the relation, `a_c = v^2/r`
⇒ `a_c = (10)^2/1000`
= `100/10^3`
= 0.1 m/s2 along RO.
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