Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
I can't solve it, what confuses me are the symbols dt and dx. I can integrate 1/x^2 in t, it's t*1/x^2, but how do I treat dx? –  Ramy Al Zuhouri Apr 15 at 12:43
1  
1/x^2 in t is t*1/x^2 ONLY if x does not depend on t, but it does. The integral of 1/x^2 is -1/x, and dt is t. That will give you a relation between x and t. –  Davidmh Apr 15 at 13:12
    
That's still not clear. I get: -1/x = At , which doesn't lead me to the correct result. –  Ramy Al Zuhouri Apr 16 at 12:57
1  
The functional solution is quite similar. You just need to put proper integration limits, and you will get your book's answer. –  Davidmh Apr 16 at 12:59

$\int^t_0 A x^2 dt = x_0 + A x^2 t$ 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, $x(t) = \dfrac{x_0}{1 - x_0At}.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.