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A train stopping at two stations $4 km$ apart takes $4 \min$ on the journey from one of the station to the other. Assuming that it first accelerates with a uniform acceleration $x$ and then that of uniform retardation $y$ , then.

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a

$\frac{1}{x} + \frac{1}{y} = 8$

b

$\frac{1}{x} + \frac{1}{y} = 4$

c

$\frac{1}{x} + \frac{1}{y} = 2$

d

$\frac{1}{x} + \frac{1}{y} = 6$

answer is C.

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Detailed Solution

A train stopping at two stations 4 km apart takes 4 min on the journey from one of the station to the other. Assuming that it first accelerates with a uniform acceleration

$\frac{v_{0}}{t_{1}} = x \Rightarrow t_{1} = \frac{v_{0}}{x} \frac{v_{0}}{t_{2}} = y \Rightarrow t_{2} = \frac{v_{0}}{y} t_{1} + t_{2} = v_{0} (\frac{1}{x} + \frac{1}{y}) = A - - - - (i)$
Further,
area = displacement
$\frac{1}{2} v_{0} \times t = s$
But numerically,

$t = s = 4 u n i t s v_{0} = 2 u n i t s$
Substituting in eq (i) we get,
$\frac{1}{x} + \frac{1}{y} = 2$

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