An automobile travelling with a speed of $60$ kmph, can apply brakes to stop within a distance of $20$ m. If the car is going twice as fast, the stopping distance will be

20

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40

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60

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80

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Solution

The correct option is D $80$ m
$u = 16.67 m / s; v = 0; s = 20; v^{2} - u^{2} = 2 a s \Rightarrow a = - 6.94 m / s^{2}$
So when $u_{1} = 2 u \Rightarrow - u_{1}^{2} = 2 a s_{1}$
Solving, $2 a s_{1} = - 4 u^{2}; s_{1} = 80 m$

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An automobile moving with a speed 60kmph can break to stop at a distance 20m. A car is moving twice as fast as this, the stopping distance will be?

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