The ‘moving walkway’ is a 300-foot long conveyor belt that moves continuously at 3 feet per ...
The ‘moving walkway’ is a 300-foot long conveyor belt that moves continuously at 3 feet per second. When Bill steps on the walkway, a group of people that are also on the walkway stands 120 feet in front of him. He walks toward the group at a combined rate (including both walkway and foot speed) of 6 feet per second, reaches the group of people, and then remains stationary until the walkway ends. What is Bill’s average rate of movement Q for his trip along the moving walkway?
Answer/Solution
5 feet per second
Steps/Work
I get E.
Their rate is 3 ft/sec and his is 6 ft/sec. Their distance is d-120 and his distance is d.
So we get two equations:
3t=d-120
6t=d
Substituting, we get 3t=6t-120, or t=40 which means that d=240 so Bill traveled 240 ft and they traveled 120 ft. (They were at 240 on the conveyor belt when they met.
So, Bill's average rate of movement Q is:
He traveled 6 ft/sec for 240 ft and 40 seconds. He traveled 3 ft/sec for the remaining 60 ft for 20 seconds.
(240+60)/(20+40)= 5 ft/sec=E
Their rate is 3 ft/sec and his is 6 ft/sec. Their distance is d-120 and his distance is d.
So we get two equations:
3t=d-120
6t=d
Substituting, we get 3t=6t-120, or t=40 which means that d=240 so Bill traveled 240 ft and they traveled 120 ft. (They were at 240 on the conveyor belt when they met.
So, Bill's average rate of movement Q is:
He traveled 6 ft/sec for 240 ft and 40 seconds. He traveled 3 ft/sec for the remaining 60 ft for 20 seconds.
(240+60)/(20+40)= 5 ft/sec=E