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A train moving with a uniform acceleration, crosses station A with a velocity of 108 kmph and another station B with a velocity of 144 kmph . The time taken for the train to travel from A to B is 20 minutes. The distance between the two stations is

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24 km

42 km

32 km

40 km

answer is B.

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Detailed Solution

$v_{1} = 108 kmph = 108 \times \frac{5}{18} = 30m/s$

$v_{2} = 144kmph = 144 \times \frac{5}{18} = 40m/s$

$t = 20 \times 60 = 12 00s$

$a = \frac{v - u}{t}$

$a = \frac{40 - 30}{1200} = \frac{1}{120} m/s$

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

$\Rightarrow {(40)}^{2} - {(30)}^{2} = 2 \times \frac{1}{120} \times s$

$\Rightarrow 700 = \frac{1}{60} \times s$

$\Rightarrow s = 42000 = 4 2km$

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