Answer:
person walk rate:
[tex]V_{p}=2.25 ft/sec[/tex]
Explanation:
A person walking on the moving sidewalk moves at a velocity:
[tex]V_{ms}=V_{p}+V_{s}[/tex]
Where [tex]V_{p} [/tex] is the velocity of the person and [tex]V_{s} [/tex] the velocity of the sidewalk.
The distance traveled in a time t is t times the velocity:
[tex]D_{ms}=V_{ms}*t=(V_{p}+V_{s})*t=65 ft[/tex]
I this same time a person on a nonmoving sidewalk travels 13ft:
[tex]D_{p}=V_{p} * t=13 ft[/tex]
Solving this for t:
[tex]t=\frac{13ft}{V_{p} }[/tex]
Replacing this on the equation for the moving sidewalk:
[tex](V_{p}+V_{s})*\frac{13ft}{V_{p}}=65 ft[/tex]
[tex]1+\frac{V_{s}}{V_{p}}=65 ft/13ft=5[/tex]
[tex]\frac{V_{s}}{V_{p}}=5-1=4[/tex]
[tex]V_{p}=\frac{V_{s} }{4}=\frac{9 ft/sec }{4}=2.25 ft/sec[/tex]
