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
Please help me^^
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
Please help me^^
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
$\displaystyle \Rightarrow t = \sqrt{\frac{m}{gk}} \ln \left( \frac{ \sqrt{mg} + \sqrt{k} v }{\sqrt{mg} - \sqrt{k} v }\right) + C$
$\displaystyle \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)?
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 $\displaystyle 0 = mg - k(50)^2$ ....