Velocity depending on position

Question

Let $v(x)$ be the velocity of a point on a line. Find the acceleration $a(x)$.

I found the following relations: $$x(t)=\int v(t) dt$$ $$v'(t)=v'(x)v(t)$$ but now I'm stuck.

Answer 1

First, let me clarify. What you are given is the velocity $v(x)$ as a function of the point's position. Velocity as a function of time is a different function so it should be denoted by something else, let's say $v_2 (t)$. Note that $x$, the position, is also a function of time. We have the following relationship:

$$v(x(t)) = v_2(t). \tag{1}$$

Now, the acceleration is the rate of change of the velocity with respect to time. Hence,

$$a_2(t) = v_2'(t).$$

This is acceleration as a function of time. Looking for $a(x)$ means you are looking for the acceleration of the point as a function of its position $x(t)$. Now, using (1), we get via chain rule,

$$a_2(t) = v'(x(t)) x'(t) = v'(x(t))v_2(t) = v'(x(t)) v(x(t)).$$

Similar to the above, we have the following relationship between $a$ and $a_2$:

$$a_2(t) = a(x(t)) = v'(x(t)) v(x(t)).$$

Looking at $a(x(t))$ as a function of $x$, we get

$$a(x) = v'(x)v(x).$$