A car travels a distance of $$x$$ with speed $$v_1$$ and then same distance $$x$$ with speed $$v_2$$ in the same direction. The average speed of the car is:

A car travels a distance $$x$$ with speed $$v_1$$ and then the same distance $$x$$ with speed $$v_2$$ in the same direction. We need to find the average speed.

Calculate the total distance.

Total distance = $$x + x = 2x$$

Calculate the total time.

Time for first part: $$t_1 = \frac{x}{v_1}$$

Time for second part: $$t_2 = \frac{x}{v_2}$$

Total time: $$t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2} = x\left(\frac{v_1 + v_2}{v_1 v_2}\right)$$

Calculate the average speed.

$$ v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2x}{x\left(\frac{v_1 + v_2}{v_1 v_2}\right)} = \frac{2v_1 v_2}{v_1 + v_2} $$

This is the harmonic mean of the two speeds, which is the correct formula when equal distances are covered at different speeds.

The correct answer is Option D: $$\frac{2v_1 v_2}{v_1 + v_2}$$.