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?
\[{\text{A}}{\text{. 4 m}}{{\text{s}}^{ - 1}}\]
${\text{B}}{\text{. 2m}}{{\text{s}}^{ - 1}}$ in the same direction of the train
${\text{C}}{\text{. 2m}}{{\text{s}}^{ - 1}}$ in the opposite direction of the train
${\text{D}}{\text{.}}$ Zero

Answer 1

Hint: The concept of relative velocity has to be used in this question in which we have to see which velocities are given in the question and which velocity needs to be found out.

Complete answer:
It is given in the question that the velocity of the train with respect to ground is constant and is 2$m{s^{ - 1}}$.
The man inside the train is moving with respect to the train with a steady speed of 2$m{s^{ - 1}}$ in the direction opposite to that in which the train is moving.
Let us assume the direction in which the train is moving to be positive and that in which the man is moving inside the train with respect to the train to be negative.
Let us represent the velocity of train with respect to ground as ${{\text{V}}_{\dfrac{{\text{T}}}{{\text{G}}}}}$ and the velocity of man with respect to ground as ${{\text{V}}_{\dfrac{{\text{M}}}{{\text{G}}}}}$
So, we can express the velocities of train as:
${{\text{V}}_{\dfrac{{\text{T}}}{{\text{G}}}}}$ = 2$m{s^{ - 1}}$
It is given in the question that the man is moving with a velocity of 2ms-1in the opposite direction as that of the train inside the train, so, he is moving with 2$m{s^{ - 1}}$ with respect to the train. Therefore,
${{\text{V}}_{\dfrac{{\text{M}}}{{\text{T}}}}}$ = -2$m{s^{ - 1}}$
So now we have to find out what will be the velocity that an observer will observe who is standing on the platform directly in front of the passenger which means we have to find out the actual velocity of the passenger who is moving inside the train.
By using the concept and formula of relative velocity we can say that:
${{\text{V}}_{\dfrac{{\text{M}}}{{\text{T}}}}}$= ${{\text{V}}_{\dfrac{{\text{M}}}{{\text{G}}}}}$- ${{\text{V}}_{\dfrac{{\text{T}}}{{\text{G}}}}}$
The given parameters are VM/T and VT/Gand we are supposed to find the value of VM.
So,
${{\text{V}}_{\dfrac{{\text{M}}}{{\text{G}}}}}$=${{\text{V}}_{\dfrac{{\text{M}}}{{\text{T}}}}}$ + ${{\text{V}}_{\dfrac{{\text{T}}}{{\text{G}}}}}$
${{\text{V}}_{\dfrac{{\text{M}}}{{\text{G}}}}}$= -2 + 2
 ${{\text{V}}_{\dfrac{{\text{M}}}{{\text{G}}}}}$= 0 ms^-1

Therefore, the velocity of the passenger that appears to the observer who is standing outside, on the platform will be zero, which means the passenger who is moving inside the train appears to be stationary to the observer.

So, the correct answer is “Option D”.

Note:
We have to carefully note in which frame the observer is present and in which frame the passenger is present and in we have to find out the velocity of the passenger in the ground frame.