# acceleration with respect to time

• Apr 8th 2008, 11:41 AM
thermalwarrior
acceleration with respect to time
Could anyone help with this?

a(t)= -g/b^2(v(t)^2 - b^2)

i need to find t in terms of v, b, g, vo,

I think I need to separate the variables but not sure how to progress.

Thanks.
• Apr 8th 2008, 12:13 PM
topsquark
Quote:

Originally Posted by thermalwarrior
Could anyone help with this?

a(t)= -g/b^2(v(t)^2 - b^2)

i need to find t in terms of v, b, g, vo,

I think I need to separate the variables but not sure how to progress.

Thanks.

Please use parenthesis! I assume this is the following:
$\frac{dv}{dt} = -\frac{g}{b^2} \cdot (v^2 - b^2)$

$\frac{dv}{v^2 - b^2} = -\frac{g}{b^2}~dt$

$\int_{v_0}^v \frac{dv}{v^2 - b^2} = -\frac{g}{b^2}\int_0^t dt$

The integral on the right is trivial. The integral on the left can be done by a partial fraction expansion:
$\int \frac{1}{v^2 - b^2} ~dv = \frac{1}{2b} \int \left ( \frac{1}{v + b} - \frac{1}{v - b} \right ) ~dv$

$= \frac{1}{2b} \cdot ln \left | \frac{v + b}{v - b} \right | + C$

So going back to the original problem we have:
$\frac{1}{2b} \cdot ln \left | \frac{v + b}{v - b} \cdot \frac{v_0 + b}{v_0 - b} \right | = \frac{g}{b^2} \cdot t$

So finally:
$t = -\frac{b}{2g} \cdot ln \left | \frac{v + b}{v - b} \cdot \frac{v_0 + b}{v_0 - b} \right |$

-Dan
• Apr 8th 2008, 12:26 PM
thermalwarrior
thats spot on, thanks!

one question- can you explain what you meant by the integral in trivial?

thanks again
• Apr 8th 2008, 01:10 PM
topsquark
Quote:

Originally Posted by thermalwarrior
thats spot on, thanks!

one question- can you explain what you meant by the integral in trivial?

thanks again

$\int dt = t + C$
That's about the simplest integral there is. Thus "trivial" as in, everyone knows it off the top of their heads kind of thing.

-Dan