# Thread: another linear motion question

1. ## another linear motion question

A particle moves vertically under gravity and a retarding force proportional to the square of the velocity. If $\displaystyle v$ is its upward or downward speed and we know that $\displaystyle \dot{v}=-g-kv^2$, show that its position at time t is given by
$\displaystyle z = z_0 + \frac{1}{k}ln\cos[\sqrtgk (t_0 -t)]$

2. What have you done so far?

3. I integrated with respect with time twice
$\displaystyle \frac{dv}{dt}=-g-kv^2$
$\displaystyle v= v_0 - gt -kv^2t$
$\displaystyle x= x_0 + v_0t -\frac{gt^2}{2} -\frac{-kv^2t^2}{2}+$

however i feel there is a technique to solce DEs that i dont know, because i cannot see where the ln cos... term would come from

4. Your technique for solving the DE is, alas, entirely wrong. The reason is that the function v occurs in the RHS. You can't just integrate with respect to t, and treat v like a constant, when it varies with t. This equation is separable. Try

$\displaystyle \displaystyle{\int\frac{dv}{-g-kv^{2}}=\int dt}.$

5. I integrate the above and get
$\displaystyle \frac{-1}{k}\int \frac{1}{v^2 + \frac{g}{k}} dv = \int dt$
$\displaystyle \frac{-1}{\sqrt (gk)}arctan (\frac{\sqrt (k)}{\sqrt(g)}v) = t - t_0$
$\displaystyle arctan(\frac{\sqrt (k)}{\sqrt (g)}v)= \sqrt (gk) (t_0 -t)$

however from here taking the tangent of both sides does not leave anything readily integrable

6. however from here taking the tangent of both sides does not leave anything readily integrable
I think it does. You basically have the tangent of a linear function of t. So the numerator looks like the derivative of the denominator. What does that give you?

7. thanks

8. You're welcome. Have a good one!