A cyclist is riding with a speed of 27kmh–1. As he approaches a circular turn on the road of radius 80m, he applies brakes and reduces his speed at the constant rate of 0.50ms–1 every second. The net acceleration of the cyclist on the circular turn is

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0.68 ms–2

0.86 ms–2

0.56 ms–2

0.76 ms–2

answer is B.

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Detailed Solution

V= 27 kmph =; r = 80mCentripetal acceleration ac = 0.703m/sec2Tangential acceleration (ac) = 0.5m/sec2

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