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 8km, and in the next minute it travels 3km. Assuming constant retardation, initial speed of the train is

The correct option is A 175ms−1s1=ut+12at2s=8kmt=1minu=?a=?⎫⎪ ⎪ ⎪⎬⎪ ⎪ ⎪⎭⇒8000=u+12a(1)2..........(1)Next minute distance, s2−s1s2=u(2)+12a(2)2=2u+2as2−s1=(2u+2a)−(u+12a)=3000=u+32a=3000..........(2)eqn(2)−eqn(1)⇒a=−5000⇒u=8000+50002=10500m/min=175m/s.

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