aditiniyer wrote:The moving walkway is a 300 foot long conveyor belt that moves continuously at 3ft/sec. When Bill steps on the walkway, a group of people that are also on the walkway stand 120 feet in front of him. He walks towards the group at a combined rate( including both walkway & foot speed) of 6 feet per second, reaches the group of people, and then remains stationary till the walkway ends. What is Bill's average rate of movement for his trip along the moving walkway ?
Ans: 5ft/sec
We are given that Bill catches up to a group of people and then remains stationary until the walkway ends. Let's determine the time, t, needed to catch up to the group of people who are 120 feet ahead of him.
Since Bill is moving at 6 ft/sec (walkway speed plus Bill's foot speed), his distance is 6t. Since the group of people is moving at 3 ft/sec (walkway speed only) and the group is 120 feet ahead, the group's distance is 3t + 120. When Bill catches up to the group, the two distances are equal. Thus, we can create the following equation and determine t:
6t = 3t + 120
3t = 120
t = 40
Therefore, it takes him 40 seconds to catch up to the group. During this 40-second period, he moves 6 x 40 = 240 feet. Since the walkway is 300 ft long, he has 300 - 240 = 60 feet remaining. However, since he is stationary after he catches up to the group, he is moving only at the walkway speed, which is 3 ft/sec. Therefore, in the last 60 feet, it takes 60/3 = 20 seconds to complete the distance, and thus it takes him a total of 40 + 20 = 60 seconds to complete the walkway from one end to another.
Since average speed = total distance/total time, his average speed = 300/60 = [spoiler]5 ft/sec.[/spoiler]
