A car is moving with a speed of 10 m/s on a circular path of radius 25m. Driver of car applies the brakes producing a uniform deceleration of $3 m / s^{2} .$ Then,

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The centripetal acceleration of car just after applying the brake is $4 m / s^{2}$

The acceleration just after applying the brake is $5 m / s^{2}$

The acceleration is directed towards the centre just at the moment of applying the brake

The angle between acceleration and velocity just after applying the brake is $127^{0}$

answer is A, B, D.

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Detailed Solution

$a_{r} = \frac{v^{2}}{r} = \frac{100}{25} = 4 m / s^{2}, a_{t} = - 3 m / s^{2}$

$∴ a = \sqrt{a_{r}^{2} + a_{t}^{2}} = 5 m / s^{2}$

Hence, $\tan θ = \frac{a_{r}}{a_{t}} = \frac{4}{3}$

$∴ θ = 53^{0}$

Thus, angle between acceleration and velocity is $180^{0} - 53^{0} = 127^{0}$

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