The motors of an electric train can give it an acceleration of $1 m s^{- 2}$ and the brakes can give a negative acceleration of $3 m s^{- 2} .$ The shortest time in which the train can make a trip between two stations $1350$ m apart is

113.6 s

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60.0 s

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245.4 s

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14.2 s

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Solution

The correct option is C 60.0 s
$a = 1 m / s^{2}$
$r = - 3 m / s^{2}$
The train accelerates for a distance $s_{1}$ and then decelerates for distance $s_{2}$ to stop at the destination.
At start, $u = 0$ ,
At $t = t_{1};$
$\Rightarrow v = u + a t_{1} = t_{1}$
$\Rightarrow s_{1} = u t + \frac{1}{2} a t_{1}^{2} = \frac{t_{1}^{2}}{2}$
At $t = t_{2};$
$\Rightarrow v = u - a t_{2} = 0$
$\Rightarrow 0 = t_{1} - 3 t_{2}$
$\Rightarrow t_{1} = 3 t_{2}$
$s_{2} = t_{1} t_{2} - \frac{1}{2} r t_{2}^{2} = \frac{t_{1}^{2}}{3} - \frac{3}{2} t_{2}^{2}$
$s_{1} + s_{2} = 1350 = \frac{t_{1}^{2}}{2} - \frac{3}{2} t_{2}^{2} + t_{1} t_{2}$
$1350 = \frac{9 t_{2}^{2}}{2} - \frac{3}{2} t_{2}^{2} + 3 t_{2}^{2}$
$\frac{1350}{6} = t_{2}^{2}$
$\Rightarrow t_{2} = 15 \Rightarrow t_{1} = 45$
$\Rightarrow t_{1} + t_{2} = 15 + 45 = 60 s e c .$

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The motors of an electric train can give it an acceleration of 1ms−2 and the brakes can give a negative acceleration of 3ms−2. The shortest time in which the train can make a trip between two stations 1350 m apart is

The motors of an electric train can give it an acceleration of $1 m s^{- 2}$ and the brakes can give a negative acceleration of $3 m s^{- 2} .$ The shortest time in which the train can make a trip between two stations $1350$ m apart is