Barry's Moving Walkway

[sgyoung12345]

Barry walks from one end to the other of a 30-meter long moving walkway at a constant rate in 30 seconds, assisted by the walkway. When he reaches the end, he reverses direction and continues walking with the same speed, but this time it takes him 120 seconds because he is traveling against the direction of the moving walkway. If the walkway were to stop moving, how many seconds would it take Barry to walk from one end of the walkway to the other?

A)48
B)60
C)72
D)75
E)80

I understand the explanation given, but I was wondering if the one I used just happened to work out of if its also a way to think about the problem.

I first calculated the rate for Barry going one way and the other and arrived at 1 meter/sec going...and 1/4 meter coming back. Given that the walkway is helping/going against Barry at the same speed, I decided to somewhat ignore and just average the 2 rates I got to get 5/4 meter/sec divided by 2, or 5/8. From there I just used that rate to get to the 48 as provided. Is this a legitimate way to think about the problem, in terms of the averaging I did?

Thanks!

[vikash.121186]

We all know that Speed * Time = Distance.
If you look at the question, the total number of steps or the distance that you have to cover either ways is a constant i.e.

Speed * Time = Distance(constant) for the given question.
which means that ratio of the speeds is equal to the inverse ratio of the time.I will elaborate the above statement with the values from the question in the later part.However, using the concept of the relative speed, it can be solves as below.

Let the speed of the man be denoted by 's'.
And the speed of the walkway or the escalator by 'e'.

When he is assisted by the walkway, his relative speed is s+e.Since, he takes 30s to cover 30-meters, we have
30*(s+e)=30 which further reduces to s+e=1;

For the situation when he reverses his direction, his speed will be reduced as then he will be walking against the direction of the motion of the walkway, which boils down to the below equation as 120*(s-e)=30 which reduces to s-e=1/4.

Adding the above two equations, one gets 2s=5/4 or s=5/8.

So, we got the speed of the man. The given distance is 30m.
Hence, the solution can be obtained by dividing the distance 30m by the man's speed(as the walkway has been stopped)which is 30/(5/8) which is equal to 48.

***However, to solve the above question, one does not need the distance which is 30m as given.

Like i said, the total distance for movement both ways is constant and as such the ratio of the time is equal to the inverse ratio of the effective speeds(or relative speed in this case) both ways which is

(s+e)/(s-e)=120/30=4.
implies, s+e=4s-4e
=>5e=3s.

The time taken to cover the distance along the direction of motion of the walkway is 30s.
Therefore, total distance is 30*(s+e).

Since, e=3s/5
=>total distance = 30*((3s/5)+s)) = 30*8s/5=48s.

As per the question, one needs to find the time taken to cover the distance when the walkway stops moving.
Hence, the solution is 48s/s=48.

Thanks & Regards

[RonPurewal]

Barry walks from one end to the other of a 30-meter long moving walkway at a constant rate in 30 seconds, assisted by the walkway. When he reaches the end, he reverses direction and continues walking with the same speed, but this time it takes him 120 seconds because he is traveling against the direction of the moving walkway. If the walkway were to stop moving, how many seconds would it take Barry to walk from one end of the walkway to the other?

A)48
B)60
C)72
D)75
E)80

I understand the explanation given, but I was wondering if the one I used just happened to work out of if its also a way to think about the problem.

I first calculated the rate for Barry going one way and the other and arrived at 1 meter/sec going...and 1/4 meter coming back. Given that the walkway is helping/going against Barry at the same speed, I decided to somewhat ignore and just average the 2 rates I got to get 5/4 meter/sec divided by 2, or 5/8. From there I just used that rate to get to the 48 as provided. Is this a legitimate way to think about the problem, in terms of the averaging I did?

Thanks!

yeah, that's beautiful.
is that not mentioned in the solution? if not, well, it certainly should be.

by the way, you aren't really "ignoring" any quantities here -- they just cancel out. if you let b stand for barry's velocity, and let w stand for the velocity of the walkway, then you have
b + w = 1
b - w = 1/4
if you add these together and divide by 2, you get b = 5/8 as described.

--

edit: i see now that the poster above me has already posted these ideas (among other things).

[wilecoyote]

Say, I'd chose 'w', the speed of the walkway and 'b' the speed of Barry. Then, selecting the direction of speed of the walkway as the positive reference, I have the following equations:

w+b=1
w-b=1/4

I have w=5/8 and b=3/8, what's wrong with that reference selection?

Thanks,

[RonPurewal]

Say, I'd chose 'w', the speed of the walkway and 'b' the speed of Barry. Then, selecting the direction of speed of the walkway as the positive reference, I have the following equations:

w+b=1
w-b=1/4

I have w=5/8 and b=3/8, what's wrong with that reference selection?

Thanks,

It's b - w = 1/4. Not w - b.
If w - b were positive, that would mean that the walkway is faster than Barry is. Which means a round trip would be impossible, because the walkway would keep carrying Barry backward when he tried to walk against it.

[alexia]

Why is this adding Barry and the walkway, when if we go by the relative rate rules (bodies moving toward each other) we should be subtracting?

[RonPurewal]

We're not talking about "bodies moving toward each other". That's an altogether different scenario, in which...
... both entities are on the ground
... both speeds are measured relative to the ground
In this sceranio, you'd subtract to find relative speed, since the speeds are originally expressed as non-relative (or as relative to the ground).

In this situation...
... the walkway is moving, with a speed measured relative to the ground
... Bob is ON the walkway
... Bob's speed is measured relative to the walkway, NOT relative to the ground.

In this problem, you actually start with a relative speed and must solve for Bob's non-relative speed (relative to the ground)!

[JayFara]

Umm, the poster of the question said he/she averaged going's speed (1 meter/sec) and coming back's speed (1/4 meter/sec) and got to 5/8 meter/sec. If we use average speed= 2xy/(x+y) for the same distances, we will get average speed= 2*1*(1/4)/(1+1/4) which is 2/5. Umm, am I missing something? I do really understand other ways that this question can be solved, but the way that the poster of this question has used is quite intriguing. I just don't get how he/she got to average speed of 5/8. Can someone shed some light, please? Thanks.

[RonPurewal]

i actually laughed out loud at the two "Umm"s.

you may want to scroll up to the third post of this thread, in which this is copiously explained (and written out in equations, too).

we are not finding an average speed for a whole journey here.
rather, we're looking at two speeds--one with the walkway, the other against it--and using those to solve for the person's walking speed on flat ground.

note that this has nothing to do with distances. those two equations (b + w = 1, b – w = 1/4) can still be written even if one of the distances is shorter than the other one.

[JayFara]

Alright, thanks.

[RonPurewal]

no problem.

[sahilk47]

We're not talking about "bodies moving toward each other". That's an altogether different scenario, in which...
... both entities are on the ground
... both speeds are measured relative to the ground
In this sceranio, you'd subtract to find relative speed, since the speeds are originally expressed as non-relative (or as relative to the ground).

In this situation...
... the walkway is moving, with a speed measured relative to the ground
... Bob is ON the walkway
... Bob's speed is measured relative to the walkway, NOT relative to the ground.

In this problem, you actually start with a relative speed and must solve for Bob's non-relative speed (relative to the ground)!

Can you explain this concept a little bit more? Here why are we doing opposite to what we calculate in case of finding relative speeds?

Thank you

[RonPurewal]

Can you explain this concept a little bit more?

what part of the existing explanation is giving you trouble?

[sahilk47]

Can you explain this concept a little bit more?

what part of the existing explanation is giving you trouble?

I think I am getting confused with the discussion on relative to ground / not relative to ground. My query is that in the case of say a moving train and a guy riding the bike, the individual speeds are subtracted to get the relative speed. But here, as I understand is the scenario wherein the guy is riding the bike but is on the train ( bizarre example I agree...), then why do we add their speeds? What is the logic behind adding in this case even though both are moving toward the same direction.

Thank you

[RonPurewal]

But here, as I understand is the scenario wherein the guy is riding the bike but is on the train ( bizarre example I agree...), then why do we add their speeds? What is the logic behind adding in this case even though both are moving toward the same direction.

imagine that you're sitting on the ground, watching a train go by, with someone riding a bike on the train.

think about how fast the cyclist passes YOU.
if you visualize the situation—literally, close your eyes and imagine what it would look like—then you should be able to see why the speeds are added.