# Retarding force solve for v

1. Oct 7, 2008

### carlee172

1. The problem statement, all variables and given/known data

A rock with mass m slides with initial velocity vo on a horizontal surface. A retarding force FR that the surface exerts on the rock is proportional to the square root of the instantaneous velocity of the rock FR=-kv1/2.
A) Find expression for the velocity of the rock as a function of time.
Express your answer in terms of the variables m, vo, k, and t.

2. Relevant equations

F=ma

v=vo +at

etc..

3. The attempt at a solution

vbasically included solving and substituting F=ma, FR=-kv1/2 and v=vo +at
in the end i got v=vo -kv1/2t/m

obviouslt this isnt right because i have v on both sides of the equation.. help

2. Oct 7, 2008

### Redbelly98

Staff Emeritus
Welcome to PF.

v=vo +at doesn't work here, because you need a constant acceleration for that equation.

F=ma is a good start. You'll also need to use some calculus to solve this one.

3. Oct 8, 2008

### carlee172

ok, so:

-k$$\sqrt{}v$$=m(dv/dt)
then integrate the first side from 0 to t and the left side from vo to v
and then solve for v, i got:
v=((-kt-2(v1/2)m)/2m)2

4. Oct 8, 2008

### carlee172

ok so i got the answer v=(4m2v0-4ktmv1/2+k2t2)/4m2
now, i have to find the integral (distance). for t, the rest are constants i guess

5. Oct 8, 2008

### Redbelly98

Staff Emeritus
One problem with that expression is that it's supposed to be an expression for v in terms of t and the other variables, i.e. given t, m, k, and v0, you should be able to calculate v.

But you have v there on the right-hand-side, so it's not really an equation for v.

Let's go back to this equation. I didn't really follow what you did after that, but I would separate the two variables, v and t, before integrating. I.e., put all the v's (including the dv term) on one side of the equation, and put dt on the other side. Then integrate.

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