Distance-velocity relation for a body accelerated with constant power

Question

A body of mass $m$ starting from rest from the origin ($x(0)=0$ and $v(0)=0$ at $t=0$) moves along the x-axis under the influence of a force $F$ that exerts a constant power $P$.

Question: How to find the relation between velocity $v(t)$ and the distance moved by the body $x(t)$?

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My understanding: I approached the problem in this manner. Since $P$ is constant over time and $P=F\cdot v$ (see e.g. this question or this), we have that

$$dP/dt=0 \quad \Rightarrow \quad d(F\cdot v)/dt=0$$

or, equivalently, $(dF/dt)\cdot v+F\cdot(dv/dt)=0$. This means that $v \cdot dF = -F \cdot dv$. Alternatively, since we want to find a distance-velocity expression, we may use the derivative of velocity with respect to position.

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Can one find the relation between velocity $v(t)$ and $x(t)$ and is it possible to do this without solving the trajectory $x(t)$?

Answer 1

Power is defined as $P = F\,v$ and since $a = \frac{F}{m}$ which means that acceleration as a function of speed is

$$a = \frac{P}{m v}$$

To integrate the above to get time and distance as a function of speed you can use the following differential relationships

$$ \begin{aligned} {\rm d}t & = \frac{1}{a(v)} {\rm d}v & {\rm d}x & = \frac{v}{a(v)} {\rm d}v \\ t(v) & = \int_0^v \frac{m v}{P} {\rm d}v & x(v) & = \int_0^v \frac{m v^2}{P} {\rm d}v \\ t(v) & = \frac{m v^2}{2 P} & x(v) & = \frac{m v^3}{3 P} \end{aligned} $$

The last equation can be found by the energy balance $P\, t = \tfrac{1}{2} m v^2$. Also combine the last two equations to get

$$\begin{aligned} v(t) & = \sqrt{ \frac{2 P t}{m} } & x(t) & = \sqrt{ \frac{8 P t^3}{9 m} } \end{aligned}$$

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To prove the differential relationships, consider ${\rm d}v = a(v) {\rm d} t$, and ${\rm d}t = \tfrac{1}{v} {\rm d}x = \tfrac{1}{a} {\rm d}v$.