A train is moving slowly on a straight track with a constant speed of 2 ms-1 passenger in that train starts walking at a steady speed of 2 ms-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

Question

A train is moving slowly on a straight track with a constant speed of 2 ms-1 passenger in that train starts walking at a steady speed of 2 ms-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) 2 ms-1 in the opposite direction of the train. (B) zero (C) 4 ms-1 (D) 2 ms-1

Tardigrade Admin · Accepted Answer

Relative velocity of passenger w.r.t. train =v text passenger -v text train =2-2=0 ∴ Relative velocity of the passenger w.r.t. the observer is zero.

$2 m s^{- 1}$ in the opposite direction of the train.

zero

$4 m s^{- 1}$

$2 m s^{- 1}$

Relative velocity of passenger w.r.t. train
$= v_{passenger} - v_{train} = 2 - 2 = 0$
$∴$ Relative velocity of the passenger w.r.t. the observer is zero.

2ms−1 in the opposite direction of the train.

zero

4ms−1

2ms−1

Relative velocity of passenger w.r.t. train =vpassenger ​−vtrain ​=2−2=0 ∴ Relative velocity of the passenger w.r.t. the observer is zero.

$2 m s^{- 1}$ in the opposite direction of the train.

$4 m s^{- 1}$

$2 m s^{- 1}$

Relative velocity of passenger w.r.t. train
$= v_{passenger} - v_{train} = 2 - 2 = 0$
$∴$ Relative velocity of the passenger w.r.t. the observer is zero.