Question

A train is moving slowly on a straight track with a constant speed of $2$ $m{s}^{-1}$. A passenger in that train starts walking at a steady speed of $2$ $m{s}^{-1}$ to the back of the train in the opposite direction of the motion of the train. So to an observer standing on the platform directly in front of that passenger, the velocity of the passenger appears to be?

• A. $4$ $m{s}^{-1}$
• B. $2$ $m{s}^{-1}$ in same direction of the train
• C. $2m{s}^{-1}$in the opposite direction of the train
• D. Zero

Answer 1

Answer: (D)

Zero

$V_{tg}=2m{s}^{-1}$

$V_{\frac{pass}{t}}=2m{s}^{-1}\left(opposite\right)$

$V_{train}-V_{ground}=2m{s}^{-1}$

$V_{pass}-V_{train}=-2m{s}^{-1}$

$V_{pass}-V_{ground}=0$

$\frac{V_{pass}}{g}=0$

Hence the passenger standing on ground will observe the passenger in rest