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Speed & Distance Problem 2

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by aditiniyer » Thu Jan 26, 2017 12:28 am
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
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by GMATGuruNY » Thu Jan 26, 2017 3:12 am
The 'moving walkway' is 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 ft/sec
(B) 2.5 ft/sec
(C) 3 ft/sec
(D) 4 ft/sec
(E) 5 ft/sec
Bill's rate alone = (combined rate for Bill and the walkway) - (rate of the walkway alone) = 6-3 = 3 feet per second.
Time for Bill to walk to the people 120 feet in front of him = d/r = 120/3 = 40 seconds.

In the 40 seconds it takes Bill to reach the group of people, the distance traveled by the walkway = r*t = 3*40 = 120 feet.

Since Bill has traveled 120 feet, and the walkway has traveled 120 feet, the remaining distance = 300-120-120 = 60 feet.
Time for the walkway alone to carry Bill the remaining distance = d/r = 60/3 = 20 seconds.

Average speed for the entire trip = (total distance)/(total time) = 300/(40+20) = 300/60 = 5 feet per second.

The correct answer is E.
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by Jay@ManhattanReview » Thu Jan 26, 2017 3:17 am
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
Hi aditiniyer,

Pl. post the complete question, with options. Sometime the options can make us change the strategy to appraoch a question.

Coming to the question.

Say Bill has to walk at the combined rate of 6 ft/sec for (120+d) ft to catch the group of peopl; where d ft is the distance the group people walked at 3 ft/sec before Bill meets them.

Thus,

Time taken by Bill = Time taken by the group

=> (120+d)/6 = d/3

=> d = 120 ft

=> Bill walked 240 ft at the rate of 6 ft/sec, taking 240/6 = 40 sec.

Bill walked at the rate of 3 ft/sec for the rest of the walkway (300-240=60 ft), taking 60/3 = 20 sec.

=> Average speed = 300/(40+20) = 300/60 = 5 ft/sec.

-Jay
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by Matt@VeritasPrep » Wed Feb 01, 2017 6:00 pm
I'd think of this way:

The group of people are moving, so Bill needs to catch them. He has to make up 120 feet, and he's moving 3 ft/sec faster than they are. From here, we get

Distance to make up = (Rate of gain) * Time

120 = 3 * T

40 = T

So Bill will catch them in 40 seconds. At that point he'll have walked 40 * 6 = 240 feet, so he'll have 60 feet left, at 3 ft/sec, for another 20 seconds.

That means it took him 40 + 20 = 60 seconds to cover 300 feet. Rate = D/T = 300/60 = 5.
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by Matt@VeritasPrep » Wed Feb 01, 2017 6:04 pm
Another way to do this that some students find more intuitive:

To see where Bill catches the group, realize that they each have to arrive at the same point after traveling for the same amount of time.

Bill is going to travel 120 + x feet at a rate of 6 ft/sec. The group is going to travel x feet at a rate of 3 ft/sec. Time = Distance / Rate, and since they travel for the same time, we can say

Bill's Time = Group's Time

Bill's D/R = Group's D/R

(120 + x)/6 = x/3

120 + x = 2x

x = 120

So the group travels 120 feet, which means Bill travels 240 feet, and from there the solution is the same.
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by Matt@VeritasPrep » Wed Feb 01, 2017 6:07 pm
One last approach would be to use the ratio of speeds. Since Bill travels twice as fast as the group does, the time it takes Bill to get to where the group started will be the same as the time it takes him to get to where the group will be when he catches them.

To get to where they started, it takes him 20 seconds. (20 * 6 = 120.)

So it will take him ANOTHER 20 seconds to catch them, for a total of 40 seconds, and from there the solution is the same as it was in my first post.
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by Jeff@TargetTestPrep » Wed Feb 08, 2017 6:40 pm
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]

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