Time in movement for forces as functions of position

Question

Let there be an object in rest of $1kg$ mass at $x=0$ and a force acted upon it which can be described by the equation

$F(x) = \frac{1}{(1-2x)^2}$

with $x$ belonging in $[0,1/2]$. I want to know in how much time it will move by $1/2 m$.

sο what I need is a way to connect $x$ with $t$ (where $t$ is time).

neuton's second law of motion states that $F=ma$ where $m$ is the mass of the object and $a$ its acceleration

so $F=ma=1a=a$ but $a= \frac{du}{dt}= \frac{du}{dx} \frac{dx}{dt}= \frac{du}{dx} u$ so $\frac{1}{(1-2x)^2} = \frac{du}{dx} u <=> \frac{dx}{(1-2x)^2} = u du <=> \frac{1}{2} \frac{1}{1-2x} = \frac{1}{2} u^2 <=> \frac{1}{1-2x} = u^2$

ok 2 things ... first of all i need a second opinion .... and secondly thats why i ask, to get the answer (+where ?? in the comments?? ) — Antonis Sk, Aug 01 '16 at 22:10
It's site policy for HW&E questions that the OP shows his own effort. You can easily edit your question to that effect. Who do you mean by1/2m? — Gert, Aug 01 '16 at 22:30
hmmmm alright .. and what do you mean by "who do you mean by 1/2m"?? — Antonis Sk, Aug 01 '16 at 22:52
@AntonisSk Please supply initial position and initial velocity as well. — velut luna, Aug 01 '16 at 23:10
You are doing well so far. Now that you have an equation for $u=\frac{dx}{dt}$ as a function $x$, the variables $x$ and $t$ are separable so you can integrate this equation easily. — sammy gerbil, Aug 01 '16 at 23:38

velut luna · Accepted Answer · 2016-08-01T23:52:06.847

0

Should be $$\int_0^x \frac{dx}{(1-2x)^2}=\int_0^u udu$$ $$\frac{1}{2}\left[\frac{1}{1-2x}\right]_0^x=\frac{u^2}{2}$$ $$u^2=\frac{1}{1-2x}-1$$ So $$\frac{1}{(1-2x)^2}=(u^2+1)^2$$

Then $$\frac{du}{dt}=\frac{1}{(1-2x)^2}=(u^2+1)^2$$ $$\int_0^u\frac{du}{(u^2+1)^2}=\int_0^t dt$$ By substitution with $u=\tan\theta$, we have $$t=\frac{1}{2}\left[\arctan u + \frac{u}{1+u^2}\right]$$ Now, when $x\to 1/2$, $u\to+\infty$

Time taken is $\pi/4$.

edited Aug 01 '16 at 23:52

answered Aug 01 '16 at 23:39

velut luna

4,004

alright but can you show me how did you get t = ... I mean I get what you are doing but i want to see it in detail – Antonis Sk Aug 02 '16 at 00:43
https://en.wikipedia.org/wiki/Equations_for_a_falling_body – velut luna Aug 02 '16 at 00:52
cant you show on your solution how you end up with t=1/2 (arctan u + u/(u^2 +1)) ??? you just jumped to the end .... can you please add the part im asking? – Antonis Sk Aug 02 '16 at 00:56
http://physics.stackexchange.com/questions/34366/time-taken-for-object-in-space-to-fall-to-earth – velut luna Aug 02 '16 at 01:12
still I didnt get the answer i wanted ... but its ok i will read the article a bit more – Antonis Sk Aug 02 '16 at 01:19

Time in movement for forces as functions of position

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