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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 braking force is increased by $12 .5%$ in the beginning find the distance that it travels before coming to rest

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195 m

270 m

240 m

210 m

answer is B.

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Detailed Solution

v^{2} - u^{2} = 2 as

$10^{2} - 30^{2} = 2 a (240)$

$10^{2} - 30^{2} = 2 \times \frac{112.5}{100} . a \times x$

$\frac{100 - 900}{- 900} = \frac{240}{112.5} \times \frac{100}{x}$

$\frac{8}{9} = \frac{240}{112.5} \times \frac{100}{x}$

$x = \frac{240}{112.5} \times \frac{100}{8} \times 9$

$= 239.625 = 240 m$

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