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

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answer is 240.

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Detailed Solution

Deceleration $(a) = \frac{v^{2} - u^{2}}{25} \Rightarrow a = \frac{800}{480} = \frac{10}{6}$
Now this deceleration is increased by 12.5%
New declaration $a = \frac{11.25}{6} \Rightarrow$ stopping distance $= \frac{{(30)}^{2}}{2 (\frac{11.25}{6})}$

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