Matty W
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not sure how to find acceleration.
i have the values for distance, start and end velocity.
S=700mm
U=0ms-1
V=990ms-1
T= ?
A= ?
what is the correct equation to use?

[Edited on 10-10-2011 by music1990]
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Matty W
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nope.. still confused
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John
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Is this engineering at uni?
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Matty W
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yeah, foundation year :/
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John
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Ian
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So -
Distance - 700mm
Starting velocity 0 m/s
End velocity 990 m/s
?
Assuming acceleration is linear (which means speed has an exponent of 2) you can use v=d/t to derive time.
Seems fairly short though? T is only a few seconds I think.
PS that -1 is to the power of -1, ie. metres x seconds^-1 = metres per second, don't mix that up when its written.
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Matty W
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yeah its meters per second to the power minus 1 or ms^-1
the question is
" a bullet travels 700mm along a rifle barrel, accelerating uniformly from rest ( u= 0ms^-1 ) to a muzzel velocity of 990ms^-1 ( v = 990ms^-1) , find the acceleration involved and the time the bullet was in the barrel
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FlaFFy_91
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If you find out the mass and force you can figure out the acceleration. Force = mass X acceleration
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Steve
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Paul J will tell you the answer based on an Xbox 360 Joypad
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Matty W
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i think ive worked it out. 700mm = 0.7m
0.7m/990ms = 0.000707070S
now i know time i can work out acceleration
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Ian
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Force and mass can be used but they're not relevant to that question.
We have enough to work it out
v=d/t
t=d/v
=700mm / 990
=0.7070 reoc.
Gets a bit funky because the speed increases exponentially with linear acceleration, not sure of that formula without checking it properly.
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Jake
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foundation
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Ian
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What's funny? Post your solutions.
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Jake
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i cant believe this is foundation engineering
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Matty SRi
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I have absolutely no idea what anything in this thread means
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Matty W
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well i worked out that 700mm is 0.7 meters ( because im measureing velocity in meters per second) so 0.7m devided by 990 ms^-1 = 0.0007070 (move the decimal place from your equation) then acceleration = end velocity - start velocity / time ( 990 /0.0007) = 1414285.714 ms^2
does that sound right?
sounds a bit too fast to me
[Edited on 10-10-2011 by music1990]
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Ian
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v1 = v0 + at
v1 = 0 + at
v1/t = a
990/0.0007070 = 1400282 m/s^2
1.4 million metres per second squared?
Might get it more accurate by interpolating the other equation.
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Ian
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quote: Originally posted by music1990
sounds a bit too fast to me
It's a bullet and it's doing nearly a km per second by the time it's travelled the length of the barrel. I would go with that
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Ben J
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One of my mates is head of physics.
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Ian
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1400142.8571429
It rounds to 1400143 as you get more accurate with the seconds.
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Matty W
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nice one. im starting to get it now
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Ian
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Did they show you integration and differentiation? That's the main concept to grasp because it shows properly the relationship between acceleration and speed.
Typically in a scenario like that acceleration is linear and speed is exponential. Falls down a bit in real situations because the bullet wouldn't continue to accelerate so you can't consider anything outside of the barrel in that one, that's why I initially thought it was quite short because it makes more sense if you use cars and things which can continue to accelerate at a constant rate throughout the test.
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Matty W
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nothing on that yet. i think its just to get a hang of using the equations for now. they may go over that in the first year
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allza
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Not done this for years but i'll try. First you ned to put them in correct units so s=0.7. Then the 2 equations you need to use are v^2 = u^2 + 2as and also s = ut+1/2at^2. If you use the first you get 990^2 = 2a*0.7. 980100=1.4a a= 700,071 ms2 You then need to use this value of a into the equation s=ut + 1/2 at^2 to work out t. I'll let you do that! I think thats correct any how.
[Edited on 10-10-2011 by allza]
[Edited on 10-10-2011 by allza]
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nibnob21
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I would use:
v^2 = u^2 + 2as
to work out the acceleration.
u disappears as it is 0 so after rearranging you are left with:
a = (v^2) / (2s)
= (990^2) / (2*0.7)
= 4900500 / 7
~ 700071.43 m/s^2
Then I would use:
v = u + at
to work out the time.
Again u disappears as it is 0, so after rearranging to make time the subject you get:
t = v / a
= 990 / (4900500 / 7)
= 7 / 4950
~ 1.41*10^-3 s
Using s = ut +1/2at^2 you can check that the values are correct. The ut term goes because u=0 so we are left with s = 1/2at^2 as follows:
1/2 * (4900500 / 7) * (7 / 4950)^2 = 0.7m
Job's a goodun 
MX5 Project Thread
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