Position dependent speed, how to compute position

Question

I can't solve a problem:

$ A= 0.5 (ms)^{-1}$, $ x_0 = 0.5 m $, $v(t)= A \cdot x^2 $, I have to compute the position at $t=3$ ($x_0$ is the initial position).

So my guess is that I should be able to compute the $x(t)$ formula by integrating $v(t)$:

$$ \int^t_0 A \cdot x^2 dt = x_0 + A \cdot x^2 \cdot t $$

So I get:

$$ x^2 \cdot At -x + x_0 = 0 $$

Which is a 2nd grade equation with a negative discriminant:

$$ \Delta(x) = (-1)^2 - 4 \cdot Atx_0 = 1 -4 \cdot 0.5 \cdot 3 \cdot 0.5 = 1-3=-2$$

My book includes just the solution, but it doesn't say how to get it. The solution is:

$$ x(t) = \frac{x_0}{1 - x_0At} $$

If I study it I get:

$$ x = \frac{x_0}{1 - x_0At} $$ $$ x \cdot \big( 1 - x_0At \big) = x_0 $$ $$ xx_0 \cdot At -x + x_0 =0 $$

Which is different from the one I got ($x^2 \cdot At -x + x_0 = 0$).

Answer 1

First problem: you say $v(t) = A x^2$, but that is a function of position, not time. Putting the definition right:

$$ v = \frac{dx}{dt} = A x^2 $$

You can regroup terms on the same variable:

$$ \frac{dx}{x^2} = A dt$$

And then do the integration:

$$ \int \frac{dx}{x^2} = \int A dt$$

This is homework, so I will leave the integral limits and the following details to you, but I think this should clarify it enough.

The key to your mistake is that you cannot simply do $\int x dt$, because $x$ is a function of $t$, but you don't know which one.

Answer 2

$\int^t_0 A x^2 dt = x_0 + A x^2 t$ is incorrect. You are assuming $x$ as a constant. $x$ is a function of time x(t).
Try $\dfrac{dx}{dt}=Ax^2 \implies \dfrac{dx}{x^2}=Adt$. Now integrate both the sides in appropriate limits.
$$\int_{x_0}^{x(t)}\dfrac{dx}{x^2}=\int_0^t Adt$$
$$\int_{x_0}^{x(t)}x^{-2}dx=\int_0^t Adt$$
$$|\dfrac{x^{-2+1}}{{-2+1}}|_{x_0}^{x(t)}=A(t-0)$$
$$|x^{-1}|_{x_0}^{x(t)}=-At$$
$$x(t)^{-1}-{x_0}^{-1}=-At$$
$$x(t)^{-1}={x_0}^{-1}+At $$ $$\dfrac{1}{x(t)}=\dfrac{1}{x_0}+At$$ $$\dfrac{1}{x(t)}=\dfrac{1+x_0At}{x_0}$$ Hence, $x(t) = \dfrac{x_0}{1 - x_0At}.$