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:
$\displaystyle \frac{dv}{dt} = -\frac{g}{b^2} \cdot (v^2 - b^2)$
$\displaystyle \frac{dv}{v^2 - b^2} = -\frac{g}{b^2}~dt$
$\displaystyle \int_{v_0}^v \frac{dv}{v^2 - b^2} = -\frac{g}{b^2}\int_0^t dt$
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