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. \(\dfrac{v_1v_2}{v_1+v_2}\)
• B. \(\dfrac{3v_1v_2}{2v_1+v_2}\)
• C. \(\dfrac{3v_1v_2}{v_1+2v_2}\)
• D. \(\dfrac{v_1+v_2}{v_1v_2}\)

Hint:

When different distances are covered with different speeds, average speed is always calculated using \[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}. \] Do not take the simple arithmetic mean of the speeds unless equal times are involved.

Answer 1

The Correct Option is B

Step 1: Express total travel time.
Let total distance = L. Time for first part (L/3 at speed \(v_1\)): \( t_1 = \frac{L}{3v_1} \). Time for second part (2L/3 at speed \(v_2\)): \( t_2 = \frac{2L}{3v_2} \).

Step 2: Apply average speed = total distance / total time.
\[ v_{avg} = \frac{L}{t_1+t_2} = \frac{L}{\frac{L}{3v_1}+\frac{2L}{3v_2}} = \frac{3v_1v_2}{v_2+2v_1} \] \[ \boxed{\dfrac{3v_1v_2}{2v_1+v_2}} \]