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