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Q.

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^–1every second. The net acceleration of the cyclist on the circular turn is

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a

0.56 ms^–2

b

0.68 ms^–2

c

0.76 ms^–2

d

0.86 ms^–2

answer is B.

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Detailed Solution

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$\begin{array}{l} V = 27 kmph = \frac{15}{2} m / \sec; r = 80 m \\ a_{C} = \frac{V^{2}}{r} = \frac{15}{2} \times \frac{15}{2} \times \frac{1}{80} = 0.703 m / s^{2} \end{array}$
Centripetal acceleration a_c = 0.703 m/sec²
Tangential acceleration (a_T) = 0.5 m/sec²
$\begin{array}{l} a = \sqrt{a_{C}^{2} + a_{T}^{2}} \\ = \sqrt{(0.703)^{2} + (0.5)^{2}} = 0.86 m / \sec^{2} \end{array}$

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