# Retarding Force on a Particle

## Homework Statement

A retarding force acts on a particle with mass m and initial velocity v0 moving on a horizontal surface; this force is proportional to the square root of the instantaneous velocity of the particle:

Fr = -k*sqrt(v)

Find an expression for the velocity of the rock as a function of time.

2. The attempt at a solution
I believe that this has to be solved through the integration of the force expression, possibly setting the right side equal to ma or m(dv/dt):

m*dv/dt = -k*sqrt(v)
dv/dt = -(k/m)*sqrt(v)

I don't know how to go about integrating this though, especially considering the v is present on both sides. I tried and this is the answer I got, though I doubt that it is correct:

v = (-3kt/m)^2 + vo

Thanks for any help!

Last edited:

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cristo
Staff Emeritus
Where does the 3 come from? Could you show your integration please?

Now that I think about it I think it should be done this way (though I'm not sure of this either because it ends with the square root of -kt/m... so unless k is negative it's not real):

dv/dt = -(k/m)*sqrt(v)
v = -(k/m)*int[sqrt(v)]dt
v = -(k/m)*[t*sqrt(v)*(1/v)]
v = -(k/m)*(t/v)
v^2 = -kt/m
v = sqrt(-kt/m) + v0

cristo
Staff Emeritus
No, that's not correct. You have a v on the right hand side, so cannot simply integrate the left hand side to give v. This is a separable equation: $$\frac{dv}{\sqrt{v}}=-\frac{k}{m}dt$$. Can you solve this?

So you would integrate both sides with respect to t? If that is the case, I know that the right side will be -kt/m, but I do not understand how to perform the left side's integral.

Oh. You integrate the left side with respect to v and the right with respect to t? So
int[v^-1/2]dv=-kt/m + c
1/2v^1/2= -kt/m + c
v=sqrt[-2kt/m+c]

Would c somehow have to do with v0 then? Would it be

v=sqrt[-2kt/m] + v0 ?