1. ## Physics/math: motion/calculus help

No, this is not homework ;-) I'm afraid it was some time ago I had one. I'm just studying on my own to cover any "holes", and I have apparently found the first one.

The problem is on motion, and under an optional section "kinematics equations derived from calculus": Given an initial velocity, and acceleration as a function of the velocity, calculate the time of a specific velocity. Or to be specific, when the velocity is half of the initial.

vi = 1.50 m/s
a = -3.00v2

That is, I think, I must find an equation relating the velocity and time. This is my feeble attempt:

a = dv / dt
dv = a dt
v = S a dt + vi ("S" is supposed to be an integral sign)

So far this feels right, since the final velocity should be the initial plus the sum of all changes of the velocity (ie. the acceleration), although it also feels very suspicious. But continuing...

v = S (-3.00v2) dt + vi
v = -3.00v2t + vi

...which is non-sense. However, blindly solving for t and putting v = vi / 2, gives an answer exactly twice of the correct answer.

Ok, I'm stuck here, and very ticked off that I cannot work it out even though it should be a simple and common calculation. I would be *very* greatful if someone, besides showing the right way to solve this, could make a guess at why I can't solve it. That is, what "hole" I need to fill in.

EDIT: I just realized that this might not belong in the "General Science" forum, sorry...
Last edited by Robert Andersson; 2005-Nov-22 at 07:44 PM. Reason: speling

2. Order of Kilopi
Join Date
Oct 2005
Posts
26,702
Your problem is that you are trying to integrate a(v) over t, which you don't have enough information to do. What you want to do is find a way to integrate something involving a(v) over v. So rewrite your:
dv = a dt
as
dv/a = dt
Now integrate both sides, where the left side is an integral that can be done when you substitute -3v^2 for a. Voila!

3. Originally Posted by Ken G
Your problem is that you are trying to integrate a(v) over t, which you don't have enough information to do.
I know. That was the only way I could think of to get t into the mix.

Now, thanks to you, Ken, I did finally solve it. The problem was probably me using too narrow thinking. This is how I worked it out.

dt = a-1 dv
t = S (-3.00v2)-1 dv (in my case, integrating to 0.5vi from vi)
t = [1/3.00v-1] (again, to 0.5vi from vi)
t = ((0.5vi)-1/3.00) - (vi-1/3.00) = 0.222 s

I want to thank you again for giving me enough hints to work it out, while forcing me to understand what I was doing in the process.

4. Order of Kilopi
Join Date
Oct 2005
Posts
26,702
You're more than welcome. I gave only hints because I could tell you had the knowhow to carry it out successfully.

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