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Standard XII

Physics

Relative Position

The maximum p...

Question

The maximum possible acceleration of a train moving on straight track is $10 m / s^{2}$ and maximum possible retardation is $5 m / s^{2}$ . Find the minimum time in which the train can complete a journey of $135$ m.

9 s

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4 s

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16 s

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25 s

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Solution

The correct option is A $9 s$
Consider the v-t graph
$t a n θ_{1} = 10 = \frac{x}{t_{0}}$
$t a n θ_{2} = 5 = \frac{x}{2 t_{0}}$
Displacement is given as the area under this graph, which is given as
$\frac{1}{2} x t_{0} + \frac{1}{2} x 2 t_{0} = \frac{3}{2} x t_{0} = \frac{3}{2} \frac{x}{t_{0}} t_{0}^{2} = \frac{3}{2} (10) t_{0}^{2}$
$= 15 t_{0}^{2} = 135 m$
$\Rightarrow t_{0}^{2} = 9 \Rightarrow t_{0} = 3 s e c$ as the maximum speed attained ( $3 \times 10 = 30$ $m / s$ )
Total time taken $= 3 t_{0} = 9 s e c$

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The maximum possible acceleration of a train moving on straight track is 10m/s2 and maximum possible retardation is 5m/s2. Find the minimum time in which the train can complete a journey of 135 m.

The correct option is A 9sConsider the v-t graphtanθ1=10=xt0tanθ2=5=x2t0Displacement is given as the area under this graph, which is given as12xt0+12x2t0=32xt0=32xt0t20=32(10)t20=15t20=135m⇒t20=9⇒t0=3sec as the maximum speed attained (3×10=30 m/s) Total time taken =3t0=9sec

The maximum possible acceleration of a train moving on straight track is $10 m / s^{2}$ and maximum possible retardation is $5 m / s^{2}$ . Find the minimum time in which the train can complete a journey of $135$ m.