Speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distances at which the two cars are stopped at the same instant is

1:1

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1:4

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1:8

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1:16

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Solution

The correct option is B 1:16
Both are given the same deceleration simultaneously and both finally stop.

Formula relevant to motion : $v^{2} = u^{2} - 2 a s$

Here $v = 0$

Then the equatio becomes,

$u^{2} = 2 a s$

For first car, $s_{1} = \frac{u^{2}}{2 a}$

For second car, $s_{2} = \frac{(4 u)^{2}}{2 a} = \frac{16 u^{2}}{2 a}$

Therefore, $\frac{s_{1}}{s_{2}} = \frac{1}{16}$

So, the ratio is $1 : 16$ .

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Speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distances at which the two cars are stopped at the same instant is

The correct option is B 1:16Both are given the same deceleration simultaneously and both finally stop.Formula relevant to motion : v2=u2−2asHere v=0Then the equatio becomes,u2=2asFor first car, s1=u22aFor second car, s2=(4u)22a=16u22aTherefore, s1s2=116So, the ratio is 1:16.