# Terminal Velocity and Bouyancy - F = kv

1. Sep 28, 2009

### ultimateman

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

A rock of mass 0.400 kg is released from the surface and sinks in the ocean. As the rock descends it is acted upon by three forces: gravity, buoyancy, and drag. The buoyancy is an upward force equal to half its weight. Drag from the water can be modeled by F = kv, where k = 0.650 kg/s.

(a) Determine the terminal speed of the sinking rock.

(b) Determine its depth, speed, and acceleration 1.50 seconds after it is released.

(c) At what depth will it be at 99.0% of its terminal speed?

2. Relevant equations

Net force (vertical), F = ma, F = kv

3. The attempt at a solution

The solution to a) was easy enough.

Fnet = mg + FB + Fdrag

Fnet = -3.92 N + 1.96 N + Fdrag = 0 N (at terminal speed)

Fdrag = 1.96 N = 0.650 kg/s *vt

vt = 3.02m/s.

b) For calculating the depth, speed, and acceleration, I think I need to integrate the function of the net force with respect to time to get the equations for velocity and position. But I have not had much luck doing so because I am rusty on my integration.

The equation for the acceleration is

a(t) = -4.90 N + 1.63s-1 *v(t)

so I'm thinking

dv/dt = -4.90 N + 1.63s-1 *dx/dt

but then I get something like

dv = -4.90 N dt + 1.63s-1 *dx

and I don't know what to do with dx?

c) I'm thinking the answer for this part will be easy enough once I have the answer to b...

Last edited: Sep 28, 2009
2. Sep 28, 2009

### ultimateman

bump

3. Sep 29, 2009

### Redbelly98

Staff Emeritus
Instead of dx/dt, try leaving that as v in this equation.