# Kinematics, find x given v.

• Dec 7th 2010, 06:30 AM
darkvelvet
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?
• Dec 7th 2010, 06:38 AM
Ackbeet

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

\$\displaystyle \dfrac{dx}{dt}=1+e^{-2t}?\$
• Dec 7th 2010, 07:05 AM
darkvelvet
\$\displaystyle \dfrac{dx}{dt}=1+e^{-2x} \$
according to the question given...
• Dec 7th 2010, 07:07 AM
Ackbeet
In that case, the equation is linear (autonomous, even), if you view t = t(x). Try that approach. What does that give you?
• Dec 7th 2010, 07:19 AM
darkvelvet
So t = integrate the inverse of \$\displaystyle 1+e^{-2x}\$ wrt x?
• Dec 7th 2010, 07:21 AM
Ackbeet
If you mean multiplicative inverse, then yes.
• Dec 7th 2010, 07:22 AM
darkvelvet
Yeaps, that's what I meant :)
• Dec 7th 2010, 07:26 AM
Ackbeet
Right. So what do you get?
• Dec 7th 2010, 07:48 AM
darkvelvet
Should I be trying with u-substitution or integration by parts? Both are giving weird answers.
• Dec 7th 2010, 07:51 AM
Ackbeet
Take a look here. Click Show steps to see how they got it.
• Dec 7th 2010, 07:55 AM
darkvelvet
Ok, at least I know that my "weird" answer is RIGHT! Thanks for the link!
Now, I just need to find x.
• Dec 7th 2010, 09:56 AM
Ackbeet
Sure. Let me know how it goes.