I've always been not so bad in mathematics, but I'm terribly bad at physics. For me, abstract concept are totally understandable, but when it come to reality, I'm lost !
So, for my job, I need to understand something.
I have an object, travelling during $t_o$ on a distance $d_o$.
It has an inital speed $v_1$, and a final speed $v_2$.
I need to find an exponential curve which fits the condition, and I'm lost in my equation, I don't know where to begin.
I try to start from the equation of the acceleration. I wan't it to be exponential, so
$$ a = e^{kt} $$
The speed curve is then the integration of the acceleration
$$ v = \frac{e^{kt}}{k}$$
I can substitute condition for finding k,
$$ v_1 = \frac{e^{k\cdot0}}{k} \implies k = \frac{1}{v_1}$$
$$ v_2 = \frac{e^{k\cdot t_0}}k \implies v_2 = v_1\cdot e^{{t_0}\cdot{v_1^{-1}}}$$
Which makes no sense. Where did I go wrong ?
My final idea is to have something like $$ t = f(d) $$ so I can find the time with the distance, and vice verse.
Best Answer
The givens: $\Delta t$, $d$, $v_1$, $v_2$. And you demand that the acceleration is exponential. Let $$ a(t)=a_0 e^{\omega t} $$ Then if we say that $v_1=0$ we have after integrating $$ v(t)=\frac{a_0}{\omega}(e^{\omega t}-1) $$ Now if we say that the initial coordinate is zero ($x_1=0$), then $$ x(t)=\frac{a_0}{\omega^2}(e^{\omega t}-1)-\frac{a_0 t}{\omega} $$ Now you say that at $t_2$, $x(t_2)=d$, and at $v(t_2)=v_2$, so $$ \frac{\omega^2 d}{a_0}+\omega t_2=e^{\omega t_2}-1 $$ and $$ \frac{\omega v_2}{a_0}=e^{\omega t_2}-1\implies \frac{\omega^2 d}{a_0}+(t_{2}-\frac{v_2}{a_0})\omega=0 $$ then $$ \omega=0,\quad \frac{v_2 -a_0 t_2}{d} \equiv \omega_0 $$ Then $$ x(t)=\frac{a_0}{\omega_{0}^2}(e^{\omega_0 t}-1)-\frac{a_0 t}{\omega_0} $$ This gives you distance in terms of time. For time, set up $$ \frac{\omega_{0}^2 x}{a_0}+1+\omega_0 t=e^{\omega_0 t}=B+At=e^{At} $$ with solution $$t(x)=\frac{-\mathcal{W}\left(-e^{-B}\right)-B}{A}=\frac{-\mathcal{W}\left(-\exp\left[-\left(\frac{\omega_{0}^2 x}{a_0}+1\right)\right]\right)-\left(\frac{\omega_{0}^2 x}{a_0}+1\right)}{\omega_0} $$ with $\mathcal{W}$ the product log function. Hopefully this one is helpful :)