Question

The speed of a moving sidewalk at an airport is 9 ​ft/sec. A person can walk 65 ft forward on the moving sidewalk in the same time it takes to walk 13 ft on a nonmoving sidewalk in the opposite direction. At what rate would a person walk on a nonmoving​ sidewalk?

Answer 1

person walk rate:

`V_(p)=2.25 ft/sec`

Step-by-step explanation:

A person walking on the moving sidewalk moves at a velocity:

`V_(ms)=V_(p)+V_(s)`

Where
`V_(p)` is the velocity of the person and
`V_(s)` the velocity of the sidewalk.

The distance traveled in a time t is t times the velocity:

`D_(ms)=V_(ms)*t=(V_(p)+V_(s))*t=65 ft`

I this same time a person on a nonmoving sidewalk travels 13ft:

`D_(p)=V_(p) * t=13 ft`

Solving this for t:

`t=(13ft)/(V_(p) )`

Replacing this on the equation for the moving sidewalk:

`(V_(p)+V_(s))*(13ft)/(V_(p))=65 ft`

`1+(V_(s))/(V_(p))=65 ft/13ft=5`

`(V_(s))/(V_(p))=5-1=4`

`V_(p)=(V_(s) )/(4)=(9 ft/sec )/(4)=2.25 ft/sec`