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IR question _Two Part Analysis

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IR question _Two Part Analysis [#permalink]

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New post 02 Jul 2012, 15:58
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Please explain me how to solve this one .


Thanks a lot !!

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Re: IR question _Two Part Analysis [#permalink]

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New post 02 Jul 2012, 17:50
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Known:
Total time 15s
Total Distance: 525m

Then, the average velocity is\(\frac{525m}{15s} = 35m/s\)

With a fixed acceleration, we have a evenly spaced set of changing velocity points over time.

Then the average of the velocities is \(\frac{{V_{15}+V_0}}{{2}}\)

Thus, what two numbers in the answer choces give us an average of 35m/s?

[Reveal] Spoiler:
\(\frac{{2+68}}{2} = 35\)

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Re: IR question _Two Part Analysis [#permalink]

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New post 13 Jul 2012, 11:48
macjas wrote:
Known:
Total time 15s
Total Distance: 525m

Then, the average velocity is\(\frac{525m}{15s} = 35m/s\)

With a fixed acceleration, we have a evenly spaced set of changing velocity points over time.

Then the average of the velocities is \(\frac{{V_{15}+V_0}}{{2}}\)

Thus, what two numbers in the answer choces give us an average of 35m/s?

[Reveal] Spoiler:
\(\frac{{2+68}}{2} = 35\)


Can you please explain how the total time is 15seconds? My question might be dumb but i don quite get it. I thought till V15 time was 15 seconds.. After the 15th second with a distance 525 how can we understand that the average time is 15seconds?

Please explain..

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Re: IR question _Two Part Analysis [#permalink]

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New post 26 Jul 2012, 03:00
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Hi Desperate123, I'm afraid I don't quite understand what you mean?

If you like, you could remove the word total and just focus on the time taken to travel 525m which would give us 35m/s. That is, 35m/s is the average velocity of the plane when it covered 525m. The graph of velocity vs time (for constant accleration) looks something like this:

Attachment:
Capture.JPG
Capture.JPG [ 16.75 KiB | Viewed 2455 times ]


So if the average velocity is 35. Then 35 will be the midpoint of the line and the y-axis end points will be t=0 and t=15. The question requires us to find the x-axis coordinates of the endpoints.

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Re: IR question _Two Part Analysis [#permalink]

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New post 27 Jul 2012, 09:28
The source for this question is from GMATPill.

http://www.gmatpill.com/gmat-practice-t ... stion/1150

This question makes use of the distance-rate-time formula.

The acceleration is constant, so we know the velocity goes from a starting speed to a final speed. The one variable we do know is distance.

Distance = rate x time

We know distance AND time.

It's just that the rate is changing - but it's changing at a constant rate of acceleration. Because of that we know there is an average speed. You can rewrite the equation as:

Distance = Avg speed x Time

Avg Speed = Distance / Time
= 525 m / 15s
= 35 m/s

OK, so that is the average speed. But we have to find the starting and final speed. OK...well the average of the starting and final speed will get us the average speed.

So, adding ("starting speed" + "final speed" ) / 2 = 35 m/s

So, "starting speed" + "final speed" = 70m/s

The only 2 possible answer choices whose sum is equal to 70m/s is 2m/s and 68m/s

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Re: IR question _Two Part Analysis [#permalink]

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New post 28 Jul 2012, 03:31
A little confused with the Question. So are they saying that the plane has traveled 525 m in 15 sec including the initial Vo?

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Re: IR question _Two Part Analysis [#permalink]

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New post 04 Aug 2012, 12:51
qweert wrote:
A little confused with the Question. So are they saying that the plane has traveled 525 m in 15 sec including the initial Vo?


they don't consider the part where the plane is taxing. The 15 sec and 525m start at the point where it starts accelerating. So in 15 sec its speed goes from v0 to v15 as it covers 525m.

I agree with you that the problem is extremely confusing because it doesn't say when the 15 sec and 525m begin. I'm not convinced it's a valid gmat problem.

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Re: IR question _Two Part Analysis   [#permalink] 04 Aug 2012, 12:51
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