# Thread: Kinematics, find x given v.

1. ## Kinematics, find x given v.

A particle moves in a line so that the velocity v m/s is given by v = 1 + e^(−2x). Find the position, x m , in terms of time t seconds ( t ≥ 0 ), given that x = 0 when t = 0.

I've tried various methods but I can't seem to get the answer
x = (1/2) ln [2e^(2t)-1].

Could the question be wrong? Any hints for my first step?
I know that I should be equating dx/dt = 1 + e^(−2x). And then, integrate both sides wrt t?

$\displaystyle \dfrac{dx}{dt}=1+e^{-2x}$ or

$\displaystyle \dfrac{dx}{dt}=1+e^{-2t}?$

3. $\displaystyle \dfrac{dx}{dt}=1+e^{-2x}$
according to the question given...

4. In that case, the equation is linear (autonomous, even), if you view t = t(x). Try that approach. What does that give you?

5. So t = integrate the inverse of $\displaystyle 1+e^{-2x}$ wrt x?

6. If you mean multiplicative inverse, then yes.

7. Yeaps, that's what I meant

8. Right. So what do you get?

9. Should I be trying with u-substitution or integration by parts? Both are giving weird answers.

10. Take a look here. Click Show steps to see how they got it.

11. Ok, at least I know that my "weird" answer is RIGHT! Thanks for the link!
Now, I just need to find x.

12. Sure. Let me know how it goes.