helpmeongmat wrote:
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 for his trip along the moving walkway?
A) 2 feet per second
B) 2.5 feet per second
C) 3 feet per second
D) 4 feet per second
E) 5 feet per second
I am confused on the rate of Bill. Is it 6 or 3? I assume it is 6 for the first 240 feet and 3 when stationary for the last 60 feet. Please help! Thanks!
Ok I did this a really strange way and long way, but I think it is correct. Since the walkway is moving the people in front of him will travel x feet. So you cannot say that the people are stationary.
I just made it so Bill is moving at 3ft per sec. and distance is 120ft. the time it takes him is 40seconds. Now since we need average speed. We go back to 6ft per sec. D=RT 6ft/sec*40sec=240feet. This is the distance he had to travel just to catch up to the group of people.
Since its AVERAGE rate. we need Total Distance/Total Time.
Total D=300ft. The first 240feet Bill travels for 40seconds. The last 60feet he travels at 3ft/sec so for the last 60 feet he travels for 20secs. (t=60/3.)
Avg R= 300ft/40sec+20sec --> Avg R= 5ft/sec
Answer E.