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.

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- Apr 8th 2008, 11:41 AMthermalwarrioracceleration 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 PMtopsquark
*Please*use parenthesis! I assume this is the following:

$\displaystyle \frac{dv}{dt} = -\frac{g}{b^2} \cdot (v^2 - b^2)$

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

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

The integral on the right is trivial. The integral on the left can be done by a partial fraction expansion:

$\displaystyle \int \frac{1}{v^2 - b^2} ~dv = \frac{1}{2b} \int \left ( \frac{1}{v + b} - \frac{1}{v - b} \right ) ~dv$

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

So going back to the original problem we have:

$\displaystyle \frac{1}{2b} \cdot ln \left | \frac{v + b}{v - b} \cdot \frac{v_0 + b}{v_0 - b} \right | = \frac{g}{b^2} \cdot t$

So finally:

$\displaystyle 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 PMthermalwarrior
thats spot on, thanks!

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

thanks again - Apr 8th 2008, 01:10 PMtopsquark