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An express train moving at 30 m/s reduces its speed to 10 m/s in a distance of 240 m. If the breaking force is increased by 12.5% in the beginning find the distance that it travels before coming to rest

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270 m

240 m

210 m

195 m

answer is B.

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Detailed Solution

deceleration = $a = \frac{v^{2} - u^{2}}{2 s} \Rightarrow a = \frac{800}{480} = \frac{10}{6}$

Now this deceleration is increased by 12.5%.

New deceleration $a = \frac{11.25}{6}$

Stopping distance= $\frac{30^{2}}{2 (\frac{11.25}{6})} = 240 m$

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