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A car moving with a speed of 50 km/hr can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr the minimum stopping distance is

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18 m

6 m

24 m

12 m

answer is C.

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Detailed Solution

v² = u² + 2as
$a = \frac{v^{2} - u^{2}}{2 s}$

As the acceleration is same in both cases,

$\frac{0^{2} - 50^{2}}{2 \times 6} = \frac{0^{2} - 100^{2}}{2 \times s}$
s = 4 × 6 = 24m

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