Question

A biker travels \( \frac{1}{3} \) of the distance \( L \) with speed \( v_1 \) and \( \frac{2}{3} \) of the distance with speed \( v_2 \). Then the average speed is:

• A. \( \frac{v_1 + v_2}{2} \)
• B. \( \frac{3 v_1 v_2}{v_1 + 2 v_2} \)
• C. \( \frac{3 v_1 v_2}{2 v_1 + v_2} \)
• D. \( \frac{v_1 + v_2}{v_1 v_2} \)

Hint:

Average Speed over Unequal Distances}
Do not use arithmetic mean for average speed when distances differ.
Use: \( v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} \)
If distance ratios are given, express time in terms of each segment and substitute.

Answer 1

The Correct Option is C

Step 1: Total distance = \( L \) The biker covers: - \( \frac{L}{3} \) at speed \( v_1 \): time \( t_1 = \frac{L}{3v_1} \) - \( \frac{2L}{3} \) at speed \( v_2 \): time \( t_2 = \frac{2L}{3v_2} \) Step 2: Total time: \[ T = t_1 + t_2 = \frac{L}{3v_1} + \frac{2L}{3v_2} \] Step 3: Average speed: \[ v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{L}{\frac{L}{3v_1} + \frac{2L}{3v_2}} = \frac{1}{\frac{1}{3v_1} + \frac{2}{3v_2}} = \frac{3 v_1 v_2}{2 v_1 + v_2} \] Conclusion: The average speed is \( \boxed{\frac{3 v_1 v_2}{2 v_1 + v_2}} \)