# Drag force with differential equations, finding max speed

• gelfand
In summary, the submarine engine provides maximum constant force ##F## to propel it through the water. 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## the maximum possible speed the submarine can reach is ##v_{max} = Ae^{-Kt}##.
gelfand

## Homework Statement

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?

## Homework Equations

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

## 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:
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.

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?

gelfand said:
Dividing through by pkpk gives
v=pF−C3e−pktpk​
v = \frac{pF - C_3 e^{-pkt}}{pk}

As the solution to the differential in terms of v

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).

## 1. What is drag force and how is it related to differential equations?

Drag force is a resistive force that acts on an object moving through a fluid, such as air or water. It is proportional to the velocity of the object and can be calculated using differential equations, specifically the Navier-Stokes equations.

## 2. How can differential equations be used to find the maximum speed of an object?

Differential equations can be used to model the motion of an object and determine its velocity as a function of time. By setting the derivative of velocity with respect to time equal to zero, the maximum speed can be found when there is no acceleration.

## 3. What factors affect the drag force on an object?

The drag force on an object is affected by the density of the fluid, the object's shape and size, and the object's velocity. Additionally, the viscosity of the fluid and the presence of obstacles can also impact the drag force.

## 4. Can drag force be reduced to increase maximum speed?

Yes, drag force can be reduced by changing the shape or size of an object, or by changing the properties of the fluid it is moving through. This can allow for an increase in maximum speed as the object experiences less resistance.

## 5. Are there other methods besides differential equations to calculate drag force and maximum speed?

Yes, there are other methods such as using empirical formulas or conducting experiments. However, differential equations provide a more precise and comprehensive understanding of the relationship between drag force and maximum speed.

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