# Velocity dependant drag

1. Oct 10, 2007

### S[e^x]=f(u)^n

1. The problem statement, all variables and given/known data
An object dragged through an unknown fluid experiences a force opposite to that of its initial velocity (Vi) that is equal to -k(v^1/2). find the equation that models its instantaneous velocity

Fn = Force Net
Ff = Frictional force
Vi = Initial Velocity
V = instantaneous velocity

2. Relevant equations
Fn=ma
Ff=-k(v^1/2)

3. The attempt at a solution

Fn=Ff
ma=-k(v^1/2)
dv/dt=(-k/m)(v^1/2)
S[dv/(v^1/2)]=(-k/m)S[dt]
2(v^1/2)=-kt/m
v=(-kt/2m)^2 + Vi

which can't be right because that would mean velocity increased as time moves positively...

i'm lost. help lol
v=

2. Oct 10, 2007

### Astronuc

Staff Emeritus
Does the object have mass m? Is another force being applied to the object?

If the force is subject only to drag, then it will deceleration in proportion to kv1/2 according to the problem as stated.

If the mass falls under gravity then the mass will decelerate or even accelerate to a constant velocity where the drag force = the force of gravity.

a = dv(t)/dt = F(t)/m, where F(t) = applied force - drag force,

and the initial condition is v(t=0) = vi/.

3. Oct 10, 2007

### S[e^x]=f(u)^n

there is no mass stated and the object is traveling horizontally and not subject to gravity.

i guess I'm having trouble deriving a velocity equation more than i am having trouble understanding the situation. i'm not even sure if the way i tried it first is the right way.

all i know for sure is that the only force acting on the point is -kv^(1/2)

could u perhaps help me find an equation for its instantaneous velocity with respect to time?

4. Oct 11, 2007

### learningphysics

you need to add a constant here:

2(v^1/2)=-kt/m + C

then

v^1/2=-kt/2m + D

v=(-kt/2m)^2 + 2(-kt/2m)D + D^2

So know solve for D using initial conditions.