KCET · Physics · Motion In One Dimension
A train is moving slowly on a straight track with a constant speed of \(2 \mathrm{~ms}^{-1}\). A passenger in that train starts walking at a steady speed of \(2 \mathrm{~ms}^{-1}\) 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 \(4 \mathrm{~ms}^{-1}\)
- B \(2 \mathrm{~ms}^{-1}\)
- C \(2 \mathrm{~ms}^{-1}\) in the opposite direction of the train
- D zero
Answer & Solution
Correct Answer
(D) zero
Step-by-step Solution
Detailed explanation
Relative velocity of passenger w.r.t. train
\(=\mathrm{v}_{\text {passenger }}-\mathrm{v}_{\text {train }}=2-2=0\)
\(\therefore\) Relative velocity of the passenger w.r.t. the observer is zero.
\(=\mathrm{v}_{\text {passenger }}-\mathrm{v}_{\text {train }}=2-2=0\)
\(\therefore\) Relative velocity of the passenger w.r.t. the observer is zero.
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