# Drag force with differential equations, finding max speed

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1. Apr 14, 2017

### gelfand

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

A submarine engine provides maximum constant force $F$ to propel it through the
water.

Assume that the magnitude of the resistive drag force of the water experienced
by the submarine is $kv$, where $k$ is the drag coefficient and $v$ is the
instantaneous speed of the boat.

When the engine is switched on to full power, the submarine starts from rest and
moves horizontally in a straight line.

If the mass of the submarine is $M$

1) Write the differential equation of motion of the submarine in terms of its
speed $v$

2) Solve the equation of motion from part (1) to find the speed $v(t)$ as a
function of time $t$

3) What is the maximum possible speed $v_{max}$ the submarine can reach with
this engine?

2. Relevant equations

Newtons second states $F = ma$. Then given force $F$ I have $F = ma$.

3. The attempt at a solution

For the drag force $-kv$ I can write this as $-Kmv$ where $K$ is some constant
such that $-Kmv = -kv$.

This gives me

$$ma = -Kmv$$

Dividing by $m$ and noting that $a = \frac{dv}{dt}$ gives

$$\frac{dv}{dt} = -Kv$$

Which is a differential equation of motion in terms of the speed $v$.

Solving this differential equation as

$$\frac{1}{v} dv = -K dt$$

Then integrate both sides for

$$\ln(v) = -K t + C_1$$

Here I can take the exponential of each side and note that $e^{x + b} = Ae^x$
where $A,b$ are constants.

So I have

$$v = Ae^{-Kt}$$

For the maximum speed I need to consider where the force of motion by the engine
is balanced by the force of the drag.

At this point I will have

$$kv_{max} = ma$$

Using the solution from part (2) we can sub for $v$ as

$$k\left( A e^{-K t} \right) = ma$$

I'm not sure how to obtain the maximum speed here though?

The expression can be rewritten as

$$\frac{kA}{ma} = e^{Kt}$$

Given $K$ is some constant this is just (where $D$ is some constant)

$$\frac{kA}{ma} = e^{t} e^{K} = De^{t}$$

So

$$\frac{kA}{Dma} = e^{t}$$

I don't think that this is correct, or even know if it makes sense.

It's suggesting that the constant $k$ multiplied by the constant $A$ divided by
the product of mass, acceleration and $D$ is equal to $e^t$.

So I'm confused

Thanks

Last edited by a moderator: Apr 14, 2017
2. Apr 14, 2017

### Ygggdrasil

The submarine engine is propelling the submarine with force F and this force is being opposed by the drag -Kmv. Therefore the net force on the submarine will be F-Kmv.

3. Apr 14, 2017

### gelfand

cheers -

so I can use that at the equilibrium I will have

$$F - kv = 0$$

And this is the net force, so these are also equal to $ma$, and can be expressed as

$$F - kv = ma$$

Note that $a = \frac{dv}{dt}$ which gives

$$F - kv = m \frac{dv}{dt}$$

From here I can divide through by mass, letting $p = \frac{1}{m}$ gives

$$pF - pkv = \frac{dv}{dt}$$

Then arranging the differential for integration gives

$$\frac{1}{pF - pdk} dv = dt$$

Integrate both sides

$$\int \frac{1}{pF - pdk} dv = \int dt$$

Here I can make the sub $u = pF - pkv$ such that $du = -(pk) dv$ and
$\frac{1}{-pk} du = dv$

Then subbing these in gives

$$-\frac{1}{pk}\int \frac{1}{u} du = \int dt$$

Solving this gives

$$-\frac{1}{pk} \ln(u) = t + C_1$$

Subbing back for $u$ gives

$$- \frac{1}{pk} \ln(pF - pkv) = t + C_1$$

Then if I wanted to solve for $v$ I can multiply both sides by $- \frac{1}{pk}$
for (using a new constant $C_2$)

$$\ln(pF - pkv) = -pkt + C_2$$

Then take exponentials as

$$pF - pkv = e^{-pkt + C_2}$$

Then note that $e^{-pkt + C_2}$ is equal to $e^{-pkt} \times C_3$. Rearranging
from this gives

$$pkv = pF - C_3 e^{-pkt}$$

Dividing through by $pk$ gives

$$v = \frac{pF - C_3 e^{-pkt}}{pk}$$

As the solution to the differential in terms of $v$

Is this approach right?

4. Apr 16, 2017

### Ygggdrasil

A good way to check the result, would be to plug it back into the original differential equation to see that m dv/dt and F - kv are indeed equal. But, the answer does look correct to me (provided you use the boundary conditions to find the value of C3 and use m instead of the placeholder variable p).