Consider a train which can accelerate with an acceleration of $20 c m / s^{2}$ and slow down with deceleration of $100 c m / s^{2}$ . Find the minimum time for the train to travel between the stations 2.7km apart.

90 s

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180 s

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160 s

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240 s

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Solution

The correct option is C 180 s
Max. Speed , $v = u + a t = 0.2 t,$

$v = 0.2 t_{1}$ ------------(1)

Dec --- $v = u - a t$

$0 = v - 1.0 t_{2}, v = t_{2}$ ----(2)

$t_{1} = \frac{t_{2}}{0.2} = 5 t_{2}$

$S = u t + \frac{1}{2} a t^{2}$

$S = 2700 m$
$2700 = 0.1 [(5 t_{2})^{2} + (t_{2})^{2} - \frac{1}{2} {t_{2}}^{2}]$

$r = 80 c m$

$3 {t_{2}}^{2} = 2700$

$t_{2} = 30 s e c, t_{1} = 150 s e c$
Total time = $t_{1} + t_{2} = 150 + 30 = 180 s e c$

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Consider a train which can accelerate with an acceleration of 20cm/s2 and slow down with deceleration of 100cm/s2. Find the minimum time for the train to travel between the stations 2.7km apart.

The correct option is C 180 sMax. Speed , v=u+at=0.2t, v=0.2t1------------(1) Dec --- v=u–at 0=v–1.0t2,v=t2----(2) t1=t20.2=5t2 S=ut+12at2 S=2700m 2700=0.1[(5t2)2+(t2)2−12t22] r=80cm 3t22=2700 t2=30sec,t1=150secTotal time = t1+t2=150+30=180sec

Consider a train which can accelerate with an acceleration of $20 c m / s^{2}$ and slow down with deceleration of $100 c m / s^{2}$ . Find the minimum time for the train to travel between the stations 2.7km apart.