JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 26)
For a train engine moving with speed of $$20 \mathrm{~ms}^{-1}$$, the driver must apply brakes at a distance of 500 $$\mathrm{m}$$ 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 $$\sqrt{x} \mathrm{~ms}^{-1}$$. The value of $$x$$ is ____________.
(Assuming same retardation is produced by brakes)
Svar
200
Forklaring
By using $3^{\text {rd }}$ equation of motion
$$ \begin{aligned} & v^2=u^2+2 a s \\\\ & (0)^2=u^2+2 a s \\\\ & u^2=-2 a s \\\\ & S=\frac{u^2}{2 a}-\frac{(20)^2}{2 \times a}=500 \\\\ & \text { acceleration of the train, } a=-\frac{400}{1000}=-0.4 \mathrm{~m} / \mathrm{sec} \end{aligned} $$
Now, if the brakes are applied at $S=250 \mathrm{~m}$ i.e. half of the distance
$$ \begin{aligned} & v^2=u^2+2 a s \\\\ & v^2=(20)^2+2(-0.4) \times 250 \\\\ & v^2=400-2 \times \frac{4}{10} \times 250 \\\\ & v^2=200 \\\\ & v=\sqrt{200} \\\\ & \text { Given } \Rightarrow v=\sqrt{x} \\\\ & \therefore x=200 \end{aligned} $$
$$ \begin{aligned} & v^2=u^2+2 a s \\\\ & (0)^2=u^2+2 a s \\\\ & u^2=-2 a s \\\\ & S=\frac{u^2}{2 a}-\frac{(20)^2}{2 \times a}=500 \\\\ & \text { acceleration of the train, } a=-\frac{400}{1000}=-0.4 \mathrm{~m} / \mathrm{sec} \end{aligned} $$
Now, if the brakes are applied at $S=250 \mathrm{~m}$ i.e. half of the distance
$$ \begin{aligned} & v^2=u^2+2 a s \\\\ & v^2=(20)^2+2(-0.4) \times 250 \\\\ & v^2=400-2 \times \frac{4}{10} \times 250 \\\\ & v^2=200 \\\\ & v=\sqrt{200} \\\\ & \text { Given } \Rightarrow v=\sqrt{x} \\\\ & \therefore x=200 \end{aligned} $$