A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathit{s}$. Then security agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$. What is the ratio of the man’s running speed to the sidewalk’s speed?

1. Time taken to move along the running side walk $2.50s$.

2. Time taken to move back along the side walk is $10s$.

The term "relative motion" refers to the comparison of the relative accelerations and velocities of two rigid bodies. A man runs along a moving sidewalk from one end to the other and back, hence it covers the same path in both cases. The man and side walk are moving with constant velocity. We are given the time taken by man to cover both distances. We can use the expression of speed and find their respective distances.

Formula:

$v=\frac{d}{t}$

The direction of both man and running sidewalk is in the same direction as a sidewalk, then the total speed is,

$V_m+V_s$.

Where,

$V_m$= speed of man

$V_s$= speed of sidewalk

The direction of both man and sidewalk is in opposite direction, and then the total speed is $V_m-V_s$.

The expression of speed is,

$v=\frac{d}{t}$

In both the cases man covers equal distance d, then the expression for$V_m+V_s$is

$V_m+V_s=\frac{d}{t_1}$ …(i)

The expression for $V_m-V_s$is

$V_m-V_s=\frac{d}{t_{.2}}$ …(ii)

From equation (i) and (ii),

$$\begin{gathered} \left(V_m+V_s\right)t_1=\left(V_m-V_s\right)t_2 \\ \left(V_m+V_m\right)2.50s=\left(V_m-V_s\right)10.0s \\ \left(V_m+V_s\right)=4\left(V_m-V_s\right) \\ 3v_m=5v_s \\ \frac{v_m}{v_s}=\frac{5}{3} \\ =1.67 \end{gathered}$$

Therefore, the of man’s running speed to the sidewalk’s speed to be$1.67$