Question

A person walks up a stationary escalation in ${{\rm{t}}_1}$ second. If he remains stationary on the escalator, then it can take him up in ${{\rm{t}}_2}$ second. If the length of the escalator is L, then
a. Determine the speed of man with respect to the escalator.
b. Determine the speed of the escalator.
c. How much time would it take him to walk up the moving escalator?

Answer 1

In the given question, there are three frames of reference. The first one being the frame of reference of the man and the second one being the frame of reference of the escalator. When the man walks on the moving escalator, then it is a relative frame of reference.

Complete step by step answer:
(a) Considering the first case in which the man is moving up a stationary escalator. The man walks a distance of L (the length of the escalator) in time ${{\rm{t}}_1}$ while the escalator remains stationary. So, the speed of man with respect to the escalator is, ${v_{{\rm{M,E}}}} = \dfrac{{\rm{L}}}{{{{\rm{t}}_1}}}$.
Therefore, the speed of man with respect to the escalator is, $\dfrac{{\rm{L}}}{{{{\rm{t}}_1}}}$.
(b) Considering the second case in which the man is stationary on the escalator, the escalator is moving. The time taken by the man to cover that distance of L (the length of the escalator) is time ${{\rm{t}}_2}$ in which the escalator covers a distance of L. So, the speed of the escalator is, ${v_{\rm{E}}} = \dfrac{{\rm{L}}}{{{{\rm{t}}_2}}}$.
Therefore, the speed of the escalator is $\dfrac{{\rm{L}}}{{{{\rm{t}}_2}}}$.
(c) Considering the third case in which the man is moving up an escalator, which is also in motion. The man walks a distance of L (the length of the escalator) in time ${{\rm{t}}_1}$ while the escalator remains stationary, and it takes time ${{\rm{t}}_2}$ for the moving escalator to move a man standing on it by a distance of L.
So, the speed of man with respect to the ground is given as,

\;\;\;\;{v_{{\rm{M,E}}}} = {v_{{\rm{M,G}}}} - {v_{\rm{E}}}\\\ {v_{{\rm{M,G}}}} = {v_{{\rm{M,E}}}} + {v_{\rm{E}}} = \dfrac{{\rm{L}}}{{{{\rm{t}}_1}}} + \dfrac{{\rm{L}}}{{{{\rm{t}}_2}}} = {\rm{L}}\left[ {\dfrac{1}{{{{\rm{t}}_1}}} + \dfrac{1}{{{{\rm{t}}_2}}}} \right]\\\ {v_{{\rm{M,G}}}} = {\rm{L}}\left[ {\dfrac{{{{\rm{t}}_1} + {{\rm{t}}_2}}}{{{{\rm{t}}_1}{{\rm{t}}_2}}}} \right] \end{array}$$ So, the time taken for the man to walk up the moving escalator is, $$\begin{array}{l} T = \dfrac{{\rm{L}}}{{{v_{{\rm{M,G}}}}}}\\\ T = \left[ {\dfrac{{{{\rm{t}}_1}{{\rm{t}}_2}}}{{{{\rm{t}}_1} + {{\rm{t}}_2}}}} \right] \end{array}$$ Therefore, the speed of man w.r.t. to the moving escalator is $$\left[ {\dfrac{{{{\rm{t}}_1}{{\rm{t}}_2}}}{{{{\rm{t}}_1} + {{\rm{t}}_2}}}} \right]$$. **Note:** There can be many frames of reference in which all the frames of reference can be in motion with respect to this frame, or either of the frames mentioned above can be at rest with respect to this new frame.