# Max speed given power and resistive forces

1. Nov 2, 2015

### Jezza

1. The problem statement, all variables and given/known data
A body of mass $M$, travelling in a straight horizontal line, is supplied with constant power $P$ and is subjected to a resistance $Mkv^2$, where $v$ is its speed and $k$ is a constant. Show that the speed of the body cannot exceed a certain value $v_m$ and find an expression for $v_m$.

2. Relevant equations

$P=\frac{dW}{dt}$
$F=\frac{dv}{dt}$

Where $W$ is work, $F$ is force, $t$ is time.

3. The attempt at a solution

First, find the force applied to the body by the power:

$P = \frac{dW}{dt} = \frac{d}{dt}\int{Fdx} = \int{F \frac{dx}{dt}} = \int{Fdv}$

We conclude by the fundamental theorem of calculus:

$F = \frac{dP}{dv}$

The resultant force on the body is therefore $F_{res} = \frac{dP}{dv} - Mkv^2 = 0$ at $v_m$.
Now we can write $\frac{dP}{dv} = Mkv^2$ which gives:

$P = \int_{v_0}^{v_m}{Mkv^2dv} = \frac{1}{3}Mk (v_m^3 - v_0^3)$

If it's ok to assume that $v_0 = 0$ (which I don't think it can be - I don't think it should depend on any initial velocity):

$v_m^3 = \frac{3P}{Mk}$.

The answer is given as $v_m^3 = \frac{P}{Mk}$.

What am I doing wrong? Also why am I having to set an initial velocity? Thanks for any help :)

2. Nov 2, 2015

### Jezza

OK I've given it some more thought. Obviously because the power is constant and at $v_m$ velocity is also constant, I can write:
$W=Fd$, which by dividing both sides by time gives $P=Fv$, giving $F=\frac{P}{v}$.

So now I can just write:

$\frac{P}{v_m}=Mkv_m^2$

Which clearly gives the desired answer. That doesn't really explain why my calculus doesn't work though...

3. Nov 2, 2015

### haruspex

You can't really do that. The middle term makes no sense - you need one more d in the numerator than in the denominator inside an integral.
In fact, at any instant, P=Fv. No need for an integral.
This is clearly not right since P is constant. You would deduce that v=0.