Question

A particle starts from rest and moves in a straight line. It travels a distance \(2L\) with uniform acceleration and then moves with a constant velocity a further distance of \(L\). Finally, it comes to rest after moving a distance of \(3L\) under uniform retardation. Then the ratio of average speed to the maximum speed \( \left( \frac{V_{avg}}{V_{m}} \right) \) of the particle is:

• A. \( \frac{6}{11} \)
• B. \( \frac{7}{11} \)
• C. \( \frac{5}{11} \)
• D. \( \frac{3}{11} \)

Hint:

For motion involving acceleration, constant velocity, and deceleration, calculate total time and use \( V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} \).

Answer 1

The Correct Option is A

Step 1: Define Motion in Three Phases
1. First Phase - The particle accelerates uniformly over distance \(2L\).
2. Second Phase - The particle moves at constant velocity over distance \(L\).
3. Third Phase - The particle decelerates uniformly over distance \(3L\) until it stops.
Step 2: Define Total Time and Maximum Speed
- Using kinematic equations, the time taken for each phase is calculated.
- The maximum velocity \( V_m \) occurs at the end of the first phase and remains constant in the second phase.
Step 3: Compute the Ratio \( \frac{V_{avg}}{V_m} \) - The average speed is given by:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}}. \] - After solving, we get: \[ \frac{V_{avg}}{V_m} = \frac{6}{11}. \] Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{\frac{6}{11}}. \]