Miks15
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| posted on 25/11/09 at 08:00 PM |
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Help with simple (mabye) maths
I dont like having to ask for maths help, but my brain just isnt working right atm!
I have an equation a=Cv^2 (a is accelaration, C a constant and v velocity)
I need to change that so i can work out the time taken for the object to travel a certain distance after being given an initial velocity! Something
tells me this should be relatively straight forward but i can quite get it right now. Could anyone shed any light on the matter. One thing i thought
was mabye you had to substitute dv/dt and ds/dt in for a and v and integrate, but its not really getting me anywhere.
Cheers guys and girls 
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JoelP
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| posted on 25/11/09 at 08:08 PM |
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a=cv^2 sounds odd to start with, but its been a long time since i did maths a level.
Beware! Bourettes is binfectious.
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matt_claydon
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| posted on 25/11/09 at 08:14 PM |
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Is you a=cv^2 referring to aerodynamic drag?
The three equations you probably need are:
v = u + at
s = ut + 0.5 * a * t^2
v^2 = u^2 + 2as
(v=final velocity, u=initial velocity, t=time, s = distance)
not sure exactly what you know and what you are looking to calculate, but with a combination of those and the drag equation you should be able to find
what you need.
[Edited on 25/11/09 by matt_claydon]
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blakep82
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| posted on 25/11/09 at 08:16 PM |
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isn't it v=at?
oops took me a while to write that  and its probably wrong
[Edited on 25/11/09 by blakep82]
hang on, initial velocity? you don't mention acceleration, so isn't it v=d/t?
[Edited on 25/11/09 by blakep82]
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twybrow
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| posted on 25/11/09 at 08:19 PM |
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Try this.....
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Miks15
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| posted on 25/11/09 at 08:27 PM |
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u cant use those formulas as they are for constant accelration,
its basically the decelaration is proportional to the velocity squared. so as the velocity decreases, so does the decelaration
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Peteff
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| posted on 25/11/09 at 08:45 PM |
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What does the pointy up arrow thing mean ? 
yours, Pete
I went into the RSPCA office the other day. It was so small you could hardly swing a cat in there.
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Miks15
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| posted on 25/11/09 at 08:47 PM |
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to the power of, so squared basically
quote:
Originally posted by Peteff
What does the pointy up arrow thing mean ?
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britishtrident
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| posted on 25/11/09 at 09:05 PM |
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Hint -> think calculus
If speed is the blank of distance with respect to time.
If acceleration is the blank of speed with respect to time.
Now back going the otherway using the above
Distance is the blank blank of acceleration with respect to time.
[Edited on 25/11/09 by britishtrident]
[I] “ What use our work, Bennet, if we cannot care for those we love? .”
― From BBC TV/Amazon's Ripper Street.
[/I]
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Miks15
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| posted on 25/11/09 at 09:43 PM |
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quote:
Originally posted by britishtrident
Hint -> think calculus
If speed is the blank of distance with respect to time.
If acceleration is the blank of speed with respect to time.
Now back going the otherway using the above
Distance is the blank blank of acceleration with respect to time.
[Edited on 25/11/09 by britishtrident]
got the first 2 blanks (im hoping its change?) cant quite work out the double blank... im being slow tonight!
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NeilP
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| posted on 25/11/09 at 10:37 PM |
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Double integral of da/dt?...
If you pay peanuts...
Mentale, yar? Yar, mentale!
Drive it like you stole it!
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Liam
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| posted on 25/11/09 at 10:49 PM |
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That's all very well, and a-level mathsy, BT, but a being a function of v is messing it up for me and my dusty brain Please
enlighten us!
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McLannahan
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| posted on 25/11/09 at 11:20 PM |
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Um... I prefer the orange smarties the most.
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Liam
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| posted on 25/11/09 at 11:33 PM |
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Sod maths - I've done an excel sheet to work it out iteratively . You got any figures you want to plug in, or is it only the proper equation
you're after - in which case... er sorry
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BenTyreman
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| posted on 25/11/09 at 11:49 PM |
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It's been a very long time since I had to do calculus, but I think that this is non-linear, impossible (or certainly beyond what I remember) to
solve by hand and should be solved using numerical integration.
Using Excel, you can evaluate accel and velocity at very small time-steps between the boundary conditions, plugging the values at time-step N into
time-step N+1.
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spidersaurus
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| posted on 26/11/09 at 02:55 AM |
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quote:
Originally posted by Miks15
quote:
Originally posted by britishtrident
Hint -> think calculus
If speed is the blank of distance with respect to time.
If acceleration is the blank of speed with respect to time.
Now back going the otherway using the above
Distance is the blank blank of acceleration with respect to time.
[Edited on 25/11/09 by britishtrident]
got the first 2 blanks (im hoping its change?) cant quite work out the double blank... im being slow tonight!
I think he is looking for
If speed is the derivative of distance with respect to time.
If acceleration is the derivative of speed with respect to time.
and double blank will be double integral?
(not that change is much difference, since derivative IS change so not saying your wrong)
looks weird to me but...
a(t) = c v(t)^2
v(t) = v_o + int(a(t))
v(t) = v_o + c*int(v(t)^2)
x(t) = x_o + int(v(t))
x(t) = x_o + int( v_o + c*int(v(t)^2))
or if v is not a function
v(t) =v_o + c*t*v^2
x(t) = x_o + v_o*t + (1/2)*c*t^2*v^2
(hope i am not wrong, like everyone else, its been a while)
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