How Do You Model Instantaneous Velocity in a Fluid with Square Root Drag?

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

Homework Statement

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

Homework Equations

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

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=

 

[Astronuc]

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 kv^1/2 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) = v_i/.

 

[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?

 

[learningphysics]

Homework Statement

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

Homework Equations

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

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

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.