Q153 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
A vehicle runs on a straight road of length 'L'. It travels half the distance with speed V and the remaining distance with speed $\dfrac{3V}{2}$. Its average speed is
- A$\dfrac{4V}{3}$
- B$\dfrac{3V}{4}$
- C$\dfrac{5V}{6}$
- Correct answerD$\dfrac{6V}{5}$
Solution
Total distance is $L$.
Time taken to travel the first half distance is $t_1 = \dfrac{L/2}{V} = \dfrac{L}{2V}$.
Time taken to travel the second half distance is $t_2 = \dfrac{L/2}{3V/2} = \dfrac{L}{3V}$.
Total time taken is $t = t_1 + t_2 = \dfrac{L}{2V} + \dfrac{L}{3V} = \dfrac{5L}{6V}$.
Average speed is $v_{avg} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{L}{\dfrac{5L}{6V}} = \dfrac{6V}{5}$.
Answer: $\dfrac{6V}{5}$
Time taken to travel the first half distance is $t_1 = \dfrac{L/2}{V} = \dfrac{L}{2V}$.
Time taken to travel the second half distance is $t_2 = \dfrac{L/2}{3V/2} = \dfrac{L}{3V}$.
Total time taken is $t = t_1 + t_2 = \dfrac{L}{2V} + \dfrac{L}{3V} = \dfrac{5L}{6V}$.
Average speed is $v_{avg} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{L}{\dfrac{5L}{6V}} = \dfrac{6V}{5}$.
Answer: $\dfrac{6V}{5}$
Hence, the correct answer is option: 'D'.