Terminal Velocity and Bouyancy - F = kv

[ultimateman]

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

The solution to a) was easy enough.

F_net = mg + F_B + F_drag

F_net = -3.92 N + 1.96 N + F_drag = 0 N (at terminal speed)

F_drag = 1.96 N = 0.650 kg/s *v_t

v_t = 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...

 

[ultimateman]

bump

 

[Redbelly98]

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

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