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  • For a train engine moving with speed of , the driver must apply brakes a

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 \mathrm{x} is ______________(Assuming same retardation is produced by brakes) 

 

Option: 1

200


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

By using \mathrm{3^{rd}}equation of motion
\begin{aligned} & \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} \\ & (0)^2=\mathrm{u}^2+2 \mathrm{as} \\ & \mathrm{u}^2=-2 \mathrm{as} \\ & \mathrm{S}=\frac{\mathrm{u}^2}{2 \mathrm{a}}-\frac{(20)^2}{2 \times \mathrm{a}}=500 \end{aligned}
acceleration of the train, a=-\frac{400}{1000}=-0.4 \mathrm{~m} / \mathrm{sec}
Now, if the brakes are applied at \mathrm{S}=250 \mathrm{~m} i.e. half of the distance
\begin{aligned} & \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} \\ & \mathrm{v}^2=(20) 2+2(-0.4) \times 250 \\ & \mathrm{v}^2=400-2 \times \frac{4}{10} \times 250 \\ & \mathrm{v}^2=200 \\ & \mathrm{v}=\sqrt{200} \end{aligned}
Given \Rightarrow \mathrm{v}=\sqrt{\mathrm{x}}
\mathrm{x}=200

Posted by

Sumit Saini

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