Question

For a train engine moving with speed of \(20\, ms ^{-1}\), the driver must apply brakes at a distance of 500 \(m\)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 \(\sqrt{x} ms ^{-1}\). The value of \(x\) is ________ . (Assuming same retardation is produced by brakes)

Hint:

Remember the equations of motion and apply them carefully, paying attention to the signs of the quantities involved.

Answer 1

Correct Answer: 200

Step 1: Calculate the Retardation
Given initial velocity \( u = 20 \, \text{m/s} \), distance \( S_1 = 500 \, \text{m} \), and final velocity \( v = 0 \). Using the third equation of motion:

\[ v^2 = u^2 + 2aS \] \[ 0 = (20)^2 + 2a(500) \] \[ 0 = 400 + 1000a \] \[ a = -\frac{400}{1000} = -0.4 \, \text{m/s}^2 \]

The negative sign indicates retardation.

Step 2: Calculate the Velocity at Half the Distance
Now, the brakes are applied at half the distance, so \( S_2 = \frac{500}{2} = 250 \, \text{m} \). The initial velocity is still \( u = 20 \, \text{m/s} \). We need to find the final velocity (\( v \)) when the train crosses the station. Using the third equation of motion:

\[ v^2 = u^2 + 2aS_2 \] \[ v^2 = (20)^2 + 2(-0.4)(250) \] \[ v^2 = 400 - 200 \] \[ v^2 = 200 \] \[ v = \sqrt{200} \, \text{m/s} \]

Step 3: Find the Value of \( x \)
The velocity is given as \( \sqrt{x} \, \text{m/s} \). We have found that \( v = \sqrt{200} \, \text{m/s} \). Therefore,

\[ x = 200 \]

Conclusion: The value of \( x \) is 200.