How do I get the velocity $v$ as a function of position $x$ from the acceleration $a$ as a function of velocity?

Question

Suppose that a particle is moving with a non-constant acceleration on the $x$ axis of $$a(v)= Av^2+Bv+C$$ ($A$, $B$ and $C$ are constants) with an initial velocity of 0 on the x axis and an initial position of 0 on the x axis. How do I calculate the velocity of the particle as a function of its position on the x axis? I thought about using the chain rule of derivatives, but it didn't work.

Answer 1

• If acceleration $a$ is only a function of speed $v$ then the following equations are used to find distance and time

  $$ \begin{aligned} x - x_0 = \int_{v_0}^v \frac{v}{a}\,{\rm d}v \\ t - t_0 = \int_{v_0}^v \frac{1}{a}\,{\rm d}v \\ \end{aligned} $$

  where $t_0$ is the initial time, $x_0$ is the initial distance, and $v_0$ is the initial speed.

• If acceleration $a$ is a function of distance $x$ only, then the following equations are used to find speed and time

  $$ \begin{aligned} \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 = \int_{x_0}^x a\,{\rm d}x \\ t - t_0 = \int_{x_0}^x \frac{1}{v}\,{\rm d}x \\ \end{aligned} $$

  where $t_0$ is the initial time, $x_0$ is the initial distance, and $v_0$ is the initial speed.

All of the above can be derived from the chain rule in calculus.

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For your case with $a = A v^2 +B v + C$ the result is

$$ \begin{aligned}t-t_{0} & =\frac{2}{\sqrt{4AC-B^{2}}}\left({\rm atan}\left(\frac{2Av+B}{\sqrt{4AC-B^{2}}}\right)-{\rm atan}\left(\frac{2Av_{0}+B}{\sqrt{4AC-B^{2}}}\right)\right)\\ x-x_{0} & =\frac{1}{2A}\ln\left(\frac{Av^{2}+Bv+C}{Av_{0}^{2}+Bv_{0}+C}\right)-\frac{B}{2A}\left(t-t_{0}\right) \end{aligned} $$

Notice the above leads to this interesting intermediate relationship

$$ \frac{a}{a_0} = \mathsf{e}^{2 A (x-x_0)+B (t-t_0) }$$

where $a_0 = A v_0^2 + B v_0 + C$

Answer 2

From $a_x = dv_x/dt$, using the chain rule, we can write: $$a_x = \frac{dv}{dx}.\frac{dx}{dt} = \frac{dv}{dx}.v$$ And then, from the equation $a = f(v)$, $$f(v) = v.\frac{dv}{dx} \rightarrow \frac{v.dv}{f(v)} = dx$$

Integrating with appropriate limits should give you the desired answer:

$$\int_0^{v_f} \frac{v.dv}{f(v)} = x_f$$

Hope this helps.