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Old 09-16-2010, 05:01 AM   #1
Qdkczrdi

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Default Anyone decent with physics?
having a brain fart, what formula would i use for the following problem

Plane must reach 37m/s to take off.

Plane starts off from 0 with a constant aceleration of 2.8 m/s^2.

How long must the runway be to reach take off speed?
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Old 09-16-2010, 05:05 AM   #2
Fausqueuego

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is that miles or meters?
meters
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Old 09-16-2010, 05:23 AM   #3
NudiJuicervich

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neeyik
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Old 09-16-2010, 06:04 AM   #4
Vikonbarius

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Let's see here, our givens are:
a = 2.8m/s^2
initial velocity = 0m/s
final velocity = 37m/s

Remember that change in velocity = a * t; and we need to know time before we can go further.

So, solve for t by doing delta v / a = t and get 13.214... seconds.

So now that we have our time, we can solve for our delta x(change in position). We do this by using the equation: delta x = (v initial)t + .5(at^2), but since initial velocity is 0, it simplifies to: delta x = .5(at^2), which equals 243.9m and then do your sig figs.

That AP Physics class I am taking is pretty insane...(just wait until you get to friction)

There are a few other ways to get this, but they all involve basically the same thing. Above was done using the v^2=2a(delta x) method
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Old 09-16-2010, 06:10 AM   #5
Qdkczrdi

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Let's see here, our givens are:
a = 2.8m/s^2
initial velocity = 0m/s
final velocity = 37m/s

Remember that change in velocity = a * t; and we need to know time before we can go further.

So, solve for t by doing delta v / a = t and get 13.214... seconds.

So now that we have our time, we can solve for our delta x(change in position). We do this by using the equation: delta x = (v initial)t + .5(at^2), but since initial velocity is 0, it simplifies to: delta x = .5(at^2), which equals 243.9m and then do your sig figs.

That AP Physics class I am taking is pretty insane...(just wait until you get to friction)

There are a few other ways to get this, but they all involve basically the same thing. Above was done using the v^2=2a(delta x) method
ah thank you. i remembered the second formula but i was confused because i didnt have time either, but you showed me how to get that. Thanks and for taking the time to explain it out, understanding it is more important then just the answer [thumbup]
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Old 09-16-2010, 07:28 AM   #6
gennickO

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Just use v^2 = u^2 +2as like IS said? With the correct numbers v^2/2a is 244.26m.

ALWAYS write u, v, a, s and t at the top of the page the write down next to each what you are given and then see how they link to you what you are looking for with equations of motion.
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Old 09-16-2010, 04:44 PM   #7
cajonnmu

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Just use v^2 = u^2 +2as like IS said? With the correct numbers v^2/2a is 244.26m.

ALWAYS write u, v, a, s and t at the top of the page the write down next to each what you are given and then see how they link to you what you are looking for with equations of motion.
Good advice - examiners like to see the working and, if you make a transcription error, you'll still get partial marks for method.
In the past, when I've been stumped or out of time, I've actually got marks for just writing down the varables, like that!
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