The acceleration of a train between two stati | vedclass

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(B) The change in velocity $\Delta v$ is equal to the area under the acceleration-time $(a-t)$ graph.Assuming the train starts from rest $(v_0 = 0)$,t

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$30$

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(B) The change in velocity $\Delta v$ is equal to the area under the acceleration-time $(a-t)$ graph.Assuming the train starts from rest $(v_0 = 0)$,the velocity at any time $t$ is given by $v(t) = \int_{0}^{t} a(t) \, dt$.The maximum speed occurs when the acceleration changes from positive to zero,which is at $t = 8\,s$.Area under the $a-t$ graph from $t = 0$ to $t = 8\,s$:Area = (Area of triangle from $0$ to $4\,s$) + (Area of rectangle from $4$ to $8\,s$)Area = $\frac{1}{2} 	imes 4\,s 	imes 5\,m/s^2 + (8\,s - 4\,s) 	imes 5\,m/s^2$Area = $10\,m/s + 20\,m/s = 30\,m/s$.Therefore,the maximum speed of the train is $30\,m/s$.

The acceleration of a train between two stations is shown in the figure. The maximum speed of the train is $............\,m/s$.

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