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It was once recorded that a car left skid marks which were 300 m in length. Assuming that the car skidded to a stop with a constant acceleration of $- 4 m s^{- 2}$ , the speed of the car before it began to skid is

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$40 \sqrt{6} m s^{- 1}$

$40 \sqrt{5} m s^{- 1}$

$20 \sqrt{6} m s^{- 1}$

$30 \sqrt{5} m s^{- 1}$

answer is C.

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Detailed Solution

Displacement s = 300m, Final velocity v = 0, Acceleration a= $- 4 m s^{- 2}$ ,

We know $v^{2} - u^{2} = 2 a s \Rightarrow 0^{2} - u^{2} = 2 (- 4) 300 \Rightarrow u = \sqrt{2400} = 20 \sqrt{6} m / s u = 20 \sqrt{6} m / s^{- 1}$

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