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

a=cv^2 sounds odd to start with, but its been a long time since i did maths a level.

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]

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]

Try this.....

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

What does the pointy up arrow thing mean ?

to the power of, so squared basically

quote:

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Originally posted by Peteff
What does the pointy up arrow thing mean ?

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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]

quote:

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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]

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Double integral of da/dt?...

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!

Um... I prefer the orange smarties the most.

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

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.

quote:

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Originally posted by Miks15

quote:

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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]

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got the first 2 blanks (im hoping its change?) cant quite work out the double blank... im being slow tonight!

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