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QJEEM23Phe1dd: JEE Main 2023 Physics - Class 11 JEEM23Phe1dd Solution: For a train engine moving with speed of , the driver must apply …

For a train engine moving with speed of 20 ms−120 \mathrm{~ms}^{-1} , the driver must apply brakes at a distance of 500 m\mathrm{m} 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 x ms−1\sqrt{x} \mathrm{~ms}^{-1} . The value of xx is ____________.

(Assuming same retardation is produced by brakes)

Step-by-Step Verified Solution

By using 3rd 3^{\text {rd }} equation of motion

amp;v2=u2+2asamp;(0)2=u2+2asamp;u2=−2asamp;S=u22a−(20)22×a=500amp; acceleration of the train, a=−4001000=−0.4 m/sec\begin{aligned}& v^2=u^2+2 a s \\\\& (0)^2=u^2+2 a s \\\\& u^2=-2 a s \\\\& S=\frac{u^2}{2 a}-\frac{(20)^2}{2 \times a}=500 \\\\& \text { acceleration of the train, } a=-\frac{400}{1000}=-0.4 \mathrm{~m} / \mathrm{sec}\end{aligned}

Now, if the brakes are applied at S=250 mS=250 \mathrm{~m} i.e. half of the distance

amp;v2=u2+2asamp;v2=(20)2+2(−0.4)×250amp;v2=400−2×410×250amp;v2=200amp;v=200amp; Given ⇒v=xamp;∴x=200\begin{aligned}& v^2=u^2+2 a s \\\\& v^2=(20)^2+2(-0.4) \times 250 \\\\& v^2=400-2 \times \frac{4}{10} \times 250 \\\\& v^2=200 \\\\& v=\sqrt{200} \\\\& \text { Given } \Rightarrow v=\sqrt{x} \\\\& \therefore x=200\end{aligned}

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