Question

A train travels from one station to another at a speed of $40\, km/hour$ and returns to the first station at a speed of $60\, km/hour$. The average speed and average velocity of the train are respectively

• A. $48 \,km \,hr^{-1}, 0$
• B. $0, 48 \,km \,hr^{-1}$
• C. $50\, km \,hr^{-1}$, $50\, km \,hr^{-1}$
• D. $50\, km \,hr^{-1}, 0$

Answer 1

Answer: (A)

Let S be the distance between two stations
Time taken by train from one station to another,$t_1 = \frac{S}{40}$ hr.
Time taken for returning, $t_2 = \frac{S}{60}$ hr.
Average speed = $\frac{Total \, distance}{Total \, time} = \frac{S+S}{\frac{S}{40} + \frac{S}{60}}$ = 48 km $hr^{-1}$
Average velocity = $\frac{Displacement}{time} = \frac{0}{time} = 0$
* Shortcut Average speed = $\frac{2\upsilon_1 \upsilon_2}{\upsilon_1 + \upsilon_2} = \frac{2(40 \times 60)}{40 + 60 } $ 48 km $hr^{-1}$