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Q.

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

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a

0.68 ms-2

b

0.86 ms-2

c

0.56 ms-2

d

0.76 ms-2

answer is B.

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Detailed Solution

Here, v = 27 km h-1 = 27 ×518ms-1 v = 152 = 7.5 ms-1 r = 80 mCentripetal acceleration, ac = v2rac = (7.5 ms-1)280 m = 0.7 ms-2Tangential acceleration, at = 0.5 ms-2Magnitude of the net acceleration isa = (ac)2+(at)2 = (0.7)2+(0.5)2 = 0.86 ms-2

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