The maximum possible acceleration of a train | vedclass

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(C) To find the minimum time,we use the velocity-time graph. The train accelerates from rest to $10\ m/s$ at $10\ m/s^2$,taking $t_1 = \frac{10}{10} =

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$15$

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(C) To find the minimum time,we use the velocity-time graph. The train accelerates from rest to $10\ m/s$ at $10\ m/s^2$,taking $t_1 = \frac{10}{10} = 1\ s$. It then decelerates from $10\ m/s$ to rest at $5\ m/s^2$,taking $t_2 = \frac{10}{5} = 2\ s$. Let $t$ be the time for which it moves at a constant speed of $10\ m/s$. The total distance covered is the area under the $v-t$ graph: $Area = \frac{1}{2} 	imes (base_1 + base_2) 	imes height = \frac{1}{2} 	imes (t + (t + 1 + 2)) 	imes 10 = 135$. Simplifying,$5(2t + 3) = 135$,so $2t + 3 = 27$,which gives $2t = 24$,or $t = 12\ s$. The total time is $T = t_1 + t + t_2 = 1 + 12 + 2 = 15\ s$.

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