# Thread: Motion in a resisting medium

1. ## Motion in a resisting medium

Hello everyone.. I'm having trouble with solving this diff equation

m. dv/dt = mg - k.v^2

Show that the solution to this differential equation is,

v = (gm/k)^1/2 x (e^ct -1 / e^ct + 1), where c = 2(kg/m)^1/2

2. Originally Posted by Modernized
Hello everyone.. I'm having trouble with solving this diff equation

m. dv/dt = mg - k.v^2

Show that the solution to this differential equation is,

v = (gm/k)^1/2 x (e^ct -1 / e^ct + 1), where c = 2(kg/m)^1/2

What have you tried? Did you at least get this far: $\frac{dt}{dv} = \frac{m}{mg - k v^2}$ ?

What do you think the next step is?

Please show the working you've done and state where you get stuck.

3. Hi! Thank you so much for replying.
This is what I've got so far,
m. dv/dt = mg - kv^2
dv/dt = g - k/m.v^2
dv/dt = -k/m ( -mg/k + v^2)

Bring it to the other side,
1 / (-mg/k + v^2) .dv = -k/m dt
Integrate both sides, however we need to use table of integrals here.
(1/2a ln | a+x / a-x | + c)
So, I got
1 / 2(mg/k)^1/2 ln | (mg/k)^1/2 +v / (mg/k)^1/2 -v | = -kt/m + c

This is what I got so far.. Im not sure how to go on from here. Thank you so much

4. Originally Posted by mr fantastic
[snip]
$\frac{dt}{dv} = \frac{m}{mg - k v^2}$ ?

[snip]
$\Rightarrow t = \sqrt{\frac{m}{gk}} \ln \left( \frac{ \sqrt{mg} + \sqrt{k} v }{\sqrt{mg} - \sqrt{k} v }\right) + C$

$\Rightarrow A e^{\sqrt{\frac{m}{gk}} t} = \frac{ \sqrt{mg} + \sqrt{k} v }{\sqrt{mg} - \sqrt{k} v }$

Is there a given boundary condition (so that A can be found)?

5. yes!
A skydiver has a terminal velocity of 50m/s before opening her parachute, g = 9.8m/s^2 and her mass is 60kg, find k and her speed after 2 and 6 seconds of free-fall.

By the way, I understand how in your first step you fliped it to make t the subject.

But.. is my way of doing it correct? Or do I have to do it like your way?..
when I find t as the subject.. isn't it harder to find v as the subject at the end?

6. Originally Posted by Modernized
yes!
A skydiver has a terminal velocity of 50m/s before opening her parachute, g = 9.8m/s^2 and her mass is 60kg, find k and her speed after 2 and 6 seconds of free-fall.

By the way, I understand how in your first step you fliped it to make t the subject.

But.. is my way of doing it correct? Or do I have to do it like your way?..
when I find t as the subject.. isn't it harder to find v as the subject at the end?
If your work is correct it will reduce to what I have done.

Were you told that v = 0 when t = 0? If not, you're probably meant to assume it. Use it to get the constant of integration. Then re-arrange the expression to make v the subject.

Since the terminal velocity is 50 m/s and you know that at terminal velocity acceleration is equal to zero, you can say $0 = mg - k(50)^2$ ....