# integral of acceleration

• Sep 29th 2010, 08:16 AM
leone
integral of acceleration
To find the velocity lets integrate acceleration with respect to time. A=-bv-(-Mv(ex)+F^ext)/( Mo - kt) were M is mass F is force... not sure how to integrate it the answer it I guess everything would be like a constant but still dont see how they got the answer i would think of moving all constants out of integral but does not see like they did it that way

v = k/b* v(ex)*[( Mo- M)/ Mo]^-k/b + Vo
• Sep 29th 2010, 10:02 AM
Ackbeet

1. What is v? Velocity? Does it depend on time? As in, is dv/dt = a?
2. What does v(ex) mean?
3. What does F^ext mean?
4. What is Mo?
5. I note that the answer you posted does not contain t. I would not expect that of an indefinite integration. Are you doing a definite integral? If so, what are the limits?
• Sep 29th 2010, 10:45 AM
leone
yea v is velocity v ex is external velocity the problem is about consider a rocket subject to a liner resistive force f=-bv but no other external forces. use equation Mv=-Mv(ex)+F^ext to show that if the rocket starts from rest and ejects mass at a constant rare k=-M then its speed is given by v=(k/b)v(ex)[1-(M/Mo)^b/k] " mo is initial mass

dose not really matter already handed in the hw but i did it wrong if thats the right answer but would like to know how i would integrate it
• Sep 29th 2010, 11:20 AM
Ackbeet
So you're saying your equation of motion is

$M\dot{v}=-M v_{\text{ex}}-bv,$ correct? Here, everything in sight is constant except $v$ and $t$, correct?
• Sep 29th 2010, 11:38 AM
leone
yea,
ask someone and they got V= exp(-bt[integral (p+f^dt)/mo-kt) exp bt)dt +c

p=-mvex
said something right side is not integrable something about near x =0
• Sep 29th 2010, 11:56 AM
Ackbeet
I can't parse your answer. You need to be much more careful with parentheses, and it wouldn't hurt to learn LaTeX. In looking at the expression

exp(-bt[integral (p+f^dt)/mo-kt) exp bt)dt +c,

I note that the "[" has no closing "]". So let's say I change that to a parenthesis:

exp(-bt(integral (p+f^dt)/mo-kt) exp bt)dt +c

I now have a problem with the "dt" not matching up in the expression with the integral sign. The integral sign is inside an exponential function, but the dt is outside? That's like writing

$\displaystyle{e^{\int f}\,dt},$ which is clearly wrong.

Communication skills are at least as important as technical skills, if not more so. Clean up your notation! And learn how to type math!

I still don't really know what DE you're trying to solve, because you haven't typed it up clearly. For example, you wrote F^ext, when you probably meant $F_{\text{ext}}.$

Don't confuse subscripts with exponents! The caret symbol "^" means exponentiation in the vast majority of cases.

Please correctly write out the DE you're trying to solve. Is this DE something that you're given? Or are you supposed to derive it from first principles?
• Sep 29th 2010, 01:11 PM
leone
well just wrote it how someone gave to me but nvm i think my professor going to post solution to hw later thxs for the help
• Sep 29th 2010, 06:11 PM
Ackbeet
Ok, you're welcome. Have a good one!