When a train is stopped by applying break it stops after travelling a distance of $50 m$ . If speed of train is doubled and same retarding force is applied then it stops after travelling a distance of:

50 m

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100 m

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200 m

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400 m

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Solution

The correct option is C $200 m$
$v^{2} = u^{2} + 2 a s_{1}$ or $0 = u^{2} - 2 a s_{1}$
$s_{1} = \frac{u^{2}}{2 a}$ ..... $(i)$
when speed of train is doubled let train travell $s_{2}$ distance
so $s_{2} = \frac{(2 u)^{2}}{2 a}$ ... $(i i)$

${g i v e n s_{1} = 50 m}$
$s_{2} = 4 \frac{u^{2}}{2 a} \Rightarrow s_{2} = 4 s_{1} = 4 \times 50 = 200 m$

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