Question

A train accelerates from rest at a constant rate 'p' for some time (attains velocity 'v') and then retards to rest at the constant rate 'q'. If the total distance covered by the train is 's', then the velocity 'v' of the train is

• A. \(\left(\left(\frac{p+q}{2pq}\right)\times s\right)^{\frac{1}{2}}\)
• B. \(\left(\left(\frac{p-q}{2pq}\right)\times s\right)^{\frac{1}{2}}\)
• C. \(\left(\left(\frac{2pq}{p+q}\right)\times s\right)^{\frac{1}{2}}\)
• D. \(\left(\left(\frac{2pq}{p-q}\right)\times s\right)^{\frac{1}{2}}\)

Hint:

Use \(v^2=2as\) separately for the speeding-up and slowing-down parts and add the two distances.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The train covers two distances: \(s_1\) while speeding up from rest to \(v\) at rate \(p\), and \(s_2\) while slowing from \(v\) to rest at rate \(q\).

Step 2: Distances
From \(v^2=2as\): \(s_1=\dfrac{v^2}{2p}\) and \(s_2=\dfrac{v^2}{2q}\).

Step 3: Calculation
Total distance \(s=s_1+s_2=\dfrac{v^2}{2}\left(\dfrac{1}{p}+\dfrac{1}{q}\right)=\dfrac{v^2(p+q)}{2pq}\).

Step 4: Result
So \(v^2=\dfrac{2pq}{p+q}\,s\), giving \(v=\left(\dfrac{2pq}{p+q}\,s\right)^{1/2}\).

Final Answer:
The velocity v equals the square root of 2pq s divided by (p + q). \[ \boxed{\text{(C) }\left(\left(\dfrac{2pq}{p+q}\right)\times s\right)^{\frac{1}{2}}} \]