How to calculate this particle's position, velocity and acceleration, each as functions of time. I'm given a particle of mass $m$, at position $x$, moving through 1-space dimenion with velocity $v=\gamma(d-x)$ for constant $\gamma, d.$
I'm also given that the particle starts from $x=0$ at $t=0$.
My question is: how do I find the velocity (and position and acceleration) as a function of time?
I know it will involve $\displaystyle v=\frac{dx}{dt}=\frac{dx}{dv}\times\frac{dv}{dt}$, but I don't know $v(t)$ as a function of $t$.
 A: Is it not the case that
$v = \dfrac{dx}{dt}? \tag{1}$
I think it is; based on this hypothesis, we have
$\dfrac{dx}{dt} = \gamma(d - x), \tag{2}$
which is easy to solve; (2) may be written
$\dfrac{1}{(d - x)} \dfrac{dx}{dt} = \gamma, \tag{3}$
or
$\dfrac{1}{(x - d)} \dfrac{dx}{dt} = - \gamma. \tag{4}$
(4) in turn may be written
$\dfrac{d \ln(x - d)}{dt} = -\gamma. \tag{5}$
We can directly integrate (5):
$\int_{t_0}^t (\dfrac{d \ln(x -d)}{dt}) dt = \int_{t_0}^t  (-\gamma) dt, \tag{6}$
yielding
$\ln(x - d) - \ln(x_0 - d) = -\gamma(t - t_0), \tag{7}$
or
$\ln(\dfrac{(x -d)}{(x_0 - d)}) = -\gamma(t - t_0), \tag{8}$
or
$x - d = (x_0 - d)e^{-\gamma(t - t_0)}, \tag{9}$
or finally
$x = (x_0 - d)e^{-\gamma(t - t_0)} + d. \tag{10}$
It is easily checked by direct differentiation that (10) satisfies (2) with initial conditions $x = x_0$ at $ t = t_0$.  The specific solution with $ x = 0$ at $t = 0$ is thus
$x = d(1 - e^{- \gamma t}), \tag{11}$
and the velocity at time $t$ is thus
$v = \dfrac{dx}{dt} = \gamma d e^{-\gamma t}; \tag{12}$
the acceleration  a is then
$a = \dfrac{dv}{dt} = -\gamma^2 d e^{- \gamma t}. \tag{13}$
We note that since the velocity $v$ is given as a function of position, the mass $m$ needn't, and indeed doesn't, enter into our calculations.  
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
