Question

When the moving sidewalk at the airport is broken, as it often seems to be, it takes you to walk $50\text{s}$from your gate to baggage claim. When it is working and you stand on the moving sidewalk the entire way, without walking, it takes$75\text{s}$ to travel the same distance. How long will it take you to travel from the gate to baggage claim if you walk while riding on the moving sidewalk?

Short answer

The time taken for the man to go from the gate to the baggage counter; if he walks while riding on the moving side walk is $30\text{s}$.

Step-by-step solution

Step 1. Introduction

Let's say the distance between the gate and the baggage counter is $x$.

Speed of the moving side walk relative to the earth is${\overrightarrow{v}}_{\text{SE}}$

Speed of walking man relative to the earth (or) sidewalk is${\overrightarrow{v}}_{\text{MS}}$

By using equation of linear motion for the walking of the man without using the moving side walk.

$x=vt={\overrightarrow{v}}_{\text{MS}}(50\text{s}){\overrightarrow{v}}_{\text{MS}}=\frac{x}{50\text{s}}\dots \dots \left(1\right)$When the man is standing stationary on the moving side walk, there's no relative velocity between the side walk and the man. Hence the man is moving at the speed of the moving side walk. Again by using the equation of linear motion for the man standing on the moving side walk.$x=vt={\overrightarrow{v}}_{\text{SE}}(75\text{s}){\overrightarrow{v}}_{\text{SE}}=\frac{x}{75\text{s}}\dots \cdots \left(2\right)$

Step 2. Explanation

When the man is walking on the moving sidewalk, the speed of the man relative to the moving side walk is ${\overrightarrow{v}}_{\text{MS}}$.

Then the speed of the man relative to the earth is,${\overrightarrow{v}}_{\text{ME}}={\overrightarrow{v}}_{\text{MS}}+{\overrightarrow{v}}_{\text{SE}}\dots \dots \left(3\right)$

Let$t$ be the time taken by man to travel from gate to baggage claim when he is on the moving sidewalk.

By using equation of linear motion for the man walking on the moving side walk from gate to baggage claim is

$x={\overrightarrow{v}}_{\text{ME}}t{\overrightarrow{v}}_{\text{ME}}=\frac{x}{t}\dots \dots \left(4\right)$

Substituting all the equations (1), (2) and (4) in equation (3), we can write

$\frac{x}{t}=\frac{x}{75\text{s}}+\frac{x}{50\text{s}}$Thus, the time by the man is

$t=\frac{(75\text{s})(50\text{s})}{(75\text{s})+(50\text{s})}=30\text{s}$

Hence, the time taken for the man to go from the gate to the baggage counter; if he walks while riding on the moving side walk is$30\text{s}$ .