Solving Integral for Falling Ball in Shampoo: dv/v

[UrbanXrisis]

say a ball falling through shampoo had a drag force of F=6.5v

the differential equation for the objects motion is:

mdv/dt=6.5v

to solve it, I get all of the v value on one side and integrate them:

dv/v=(6.5/m)(dt)
integral[dv/v]=integral[(6.5/m)(dt)]
I know that integral[(6.5/m)(dt)] becomes 6.5t/m+c, but what will happen to the dv/v?

 

[Leong]

How about mg ?

 

[UrbanXrisis]

ignoring the effects of gravity, sry

 

[teclo]

say a ball falling through shampoo had a drag force of F=6.5v

the differential equation for the objects motion is:

mdv/dt=6.5v

to solve it, I get all of the v value on one side and integrate them:

dv/v=(6.5/m)(dt)
integral[dv/v]=integral[(6.5/m)(dt)]
I know that integral[(6.5/m)(dt)] becomes 6.5t/m+c, but what will happen to the dv/v?

the indefinite integral of dx/x is ln|x|+C

i guess if you are looking at this as a seperable differential equation you'd have

ln|v|=(6t/m)+C
|v|=e^((6t/m)+C)
|v|=e^(6t/m) * e^C
if K = +,- e^C
v=Ke^(6t/m)

i don't know, is that what you were looking for?

 

[Leong]

$$\int \frac{1}{v}dv=ln |v|$$