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

Let's determine the value of \(x\) using kinematics equations. Given:

• Initial speed of the train, \(u = 20\, \text{ms}^{-1}\).
• Distance to stop, \(s_1 = 500\, m\).
• Final speed at the station should be \(v = 0\, \text{ms}^{-1}\).

Step 1: Calculate deceleration \(a\) using the equation:

\(v^2 = u^2 + 2as_1\)
Substitute: \(0 = 20^2 + 2a \times 500\)
This gives: \(0 = 400 + 1000a\)
Solve for \(a\): \(a = -0.4\, \text{ms}^{-2}\) (deceleration)

Step 2: Determine speed when brakes are applied at \(\frac{500}{2} = 250\, m\).

Use the same kinematic equation for this shorter distance:

\(v^2 = u^2 + 2as_2\)
Where \(s_2 = 250\, m\).

Substitute: \(v^2 = 20^2 + 2(-0.4)(250)\)
This simplifies to: \(v^2 = 400 - 200\)
Therefore, \(v^2 = 200\)

Conclusion: The speed at which the train engine would cross the station is \(\sqrt{200}\, \text{ms}^{-1}\), which implies \(x = 200\). The value \(x\) is thus correctly calculated as 200, fitting the range [200, 200].