A train is running at full speed when brakes are applied. In the first minute it travels $8 k m$ , and in the next minute it travels $3 k m$ . Assuming constant retardation, initial speed of the train is

150 m s^{- 1}

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175 m s^{- 1}

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200 m s^{- 1}

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225 m s^{- 1}

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Solution

The correct option is A $175 m s^{- 1}$

$s_{1} = u t + \frac{1}{2} a t^{2}$
$\begin{matrix} s = 8 k m t = 1 m i n u = ? a = ? \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭$
$\Rightarrow 8000 = u + \frac{1}{2} a (1)^{2} . . . . . . . . . . (1)$
Next minute distance, $s_{2} - s_{1}$
$s_{2} = u (2) + \frac{1}{2} a (2)^{2} = 2 u + 2 a$
$s_{2} - s_{1} = (2 u + 2 a) - (u + \frac{1}{2} a) = 3000$
$= u + \frac{3}{2} a = 3000 . . . . . . . . . . (2)$
$e q^{n} (2) - e q^{n} (1) \Rightarrow a = - 5000$
$\Rightarrow u = 8000 + \frac{5000}{2} = 10500 m / m i n$
$= 175 m / s .$

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A train is running at full speed when brakes are applied. In the first minute it travels $8 k m$ , and in the next minute it travels $3 k m$ . Assuming constant retardation, initial speed of the train is