Kinematics

JEE Main level, numerical answer, medium

48 of 308

Question 48 of 308

For a train engine moving with speed of 20ms–120{ms}^{–1}, the driver must apply brakes at a distance of 500m500m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed xms−1\sqrt{x}{ms}^{-1}. The value of xx is ______. (Assuming same retardation is produced by brakes)

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Answer: 200

Given that, the initial velocity u=20m/su=20m/s, the distance of the station S1=500m{S}_{1}=500m, and the final velocity v=0v=0.
The relation between the distance travelled (S)\left(S\right) by the train with its initial velocity (u)\left(u\right) and its final velocity (v)\left(v\right) is given by
v2=u2−2aS........................(1){v}^{2}={u}^{2}-2aS........................(1)
where, aa indicates the retardation of the train.
Substitute the values corresponding to the initial condition into equation (1) and solve to calculate the retardation of the train.
0=(20)2−2×a×500a=0.4m/s2\begin{matrix}0 & = & (20{)}^{2}-2\times a\times 500 \\ a & = & 0.4m/{s}^{2}\end{matrix}
For the second scenario, the initial velocity u=20m/su=20m/s and the distance travelled by the train S2=250m{S}_{2}=250m.
Use these values along with the value of the retardation into equation (1) and solve to calculate the final velocity of the train.
v2=(20)2−2×0.4×250=200v=200..........................(2)\begin{matrix}{v}^{2} & = & (20{)}^{2}-2\times 0.4\times 250 \\ & = & 200 \\ v & = & \sqrt{200}..........................(2)\end{matrix}
Comparing the value of the final velocity obtained in equation (2) with the given expression, we get x=200x=200.

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