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A car is moving with a speed of 30 ms^–1 on a circular path of radius 500 m. If its speed is increasing at the rate of 2 ms^–2, the net acceleration of the car is

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2.7 ms^–2

2 ms^–2

1.8 ms^–2

3.6 ms^–2

answer is B.

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Detailed Solution

$\begin{array}{l} a = \sqrt{a_{r}^{2} + a_{t}^{2}} & a_{t} = 2 m / \sec^{2} \\ a_{r} = \frac{V^{2}}{r} = \frac{30 \times 30}{500} = \frac{9}{5} = 1.8 m / \sec^{2} \\ a = \sqrt{2^{2} + (1.8)^{2}} ≃ 2.7 {ms}^{- 2} \end{array}$

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