Question

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. $\frac{v_1 v_2}{2\left(v_1+v_2\right)}$
• B. $\frac{2 x}{v_1+v_2}$
• C. $\frac{2 v_1 v_2}{v_1+v_2}$
• D. $\frac{v_1+v_2}{2}$

Answer 1

The Correct Option is C

1. Total distance traveled: \[ d = x + x = 2x. \]
2. Total time taken: \[ t = \frac{x}{V_1} + \frac{x}{V_2}. \]
3. Average speed: \[ v_{\text{avg}} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2x}{\frac{x}{V_1} + \frac{x}{V_2}}. \]
4. Simplifying: \[ v_{\text{avg}} = \frac{2x}{x \left(\frac{1}{V_1} + \frac{1}{V_2}\right)} = \frac{2}{\frac{1}{V_1} + \frac{1}{V_2}}. \]
\[ v_{\text{avg}} = \frac{2 V_1 V_2}{V_1 + V_2}. \]
Thus, the average speed is \(\frac{2 V_1 V_2}{V_1 + V_2}\). The average speed for equal distances depends on the harmonic mean of the two speeds. This is because the time taken varies inversely with speed.