# Submarine Buoyancy Differential Equation

BJL13

## Homework Statement

A submarine of mass 80 000 kg is floating at rest (neutrally buoyant) at a depth of 200 m in sea water. It starts pumping out sea water from its ballast tanks at a rate of 600 litres per minute, thus affecting both its mass and the buoyancy force. Determine the vertical velocity of the submarine after 10 seconds of ascent, assuming that the ballast tanks are large enough to be emptied at a constant rate throughout the ascent. [You may assume that the density of sea water is 1 kg per litre. The acceleration due to gravity g =10 m s−2 . Also assume that water pumped from the tanks leaves the submarine at negligible velocity, and the air in the empty part of the tank has negligible weight.]

## Homework Equations

Archimedes Principle
Newton's Second Law

## The Attempt at a Solution

Let V = volume of submarine = 80000m^3 (from equal densities at t=0)
From N2L applied upwards
m(t) * a = Vg-m(t)g
So I think a = Vg/m(t) -g

m(t)=80000-dm/dt * t
dm/dt = 600/60 =10
m(t) = 80000-10t

Plugging this in leads to a = 8000g/(8000-t) - g

Is this correct?

Thank you very much in advance...

I note that the problem doesn't mention the effect of drag due to the viscosity of the water 