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.

  • $\begingroup$ You are just not using suitable integration constants. $\endgroup$ – Yaman Sanghavi Jun 23 '17 at 10:02

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 :)


Let's call the starting and stopping points $x_1$ and $x_2$, the times $t_1,t_2$ and speeds $v_1, v_2$. Let's for easier usage set $t_1 = 0$.

Assuming the exponential acceleration $a(t)= A\exp (kt)$, $[k]=\tfrac{1}{s},[A]=\tfrac{m}{s^2} $. let formally $A=1\tfrac{m}{s^2}$

speed becomes (note the constant) $$ v(t) = C + \frac{A}{k}\exp(kt)\\ v(0) = v_1 = C + \frac{A}{k} \\ \Rightarrow C = v_1-\frac{A}{k} $$ so your second condition is: $$ v(t_2) = v_2 = v_1 + \frac{A}{k}(\exp(kt_2)-1)\\ v_2-v_1 +\frac{A}{k} = A\exp( -\ln(k) k t_2)\\ -\frac{ln(v_2-v_1+\tfrac{A}{k})-\ln{A}}{t_2} = const = \ln(k) k $$

from here one should solve it numerically. Even better approach is suggested by Emilio Pisanty in comments.

distance is then: $$ x(t) = x_1 + (v_1-\tfrac{A}{k}) t + \frac{A}{\hat{k}^2}\exp(\hat{k}t)\\ $$

  • $\begingroup$ Something must me goofy because the last equation doesn't have the right units... $\endgroup$ – kηives Jun 6 '12 at 16:13
  • $\begingroup$ true, goes right from the $\exp$ assumption, let's introduce a factor $\endgroup$ – IljaBek Jun 6 '12 at 16:17
  • $\begingroup$ You need to set $v(t)=v_1+{A\over k}\left(\exp(kt)-1\right)$ or it won't obey $v(0)=v_1$. One then needs to solve numerically $\frac{v_2-v_1}{At_2}=\frac{e^{kt_2}-1}{kt_2}$, which is suitably in terms of dimensionless quantities. $\endgroup$ – Emilio Pisanty Jun 6 '12 at 18:09
  • 2
    $\begingroup$ The numerical solution can also be avoided using tables for the product log function: wolframalpha.com/input/?i=%28e%5Ex-1%29%2Fx+inverse $\endgroup$ – Emilio Pisanty Jun 6 '12 at 18:13
  • $\begingroup$ @EmilioPisanty: correcting, thx. good tip, but numerical solution stays numerical ;) $\endgroup$ – IljaBek Jun 6 '12 at 21:56

The problem is that exponential acceleration means damping, and this requires one parameter, yet two constraints exists (final speed and distance).

The simplest form would be $a(t) = C_1 v(t)$ which solved by

$$v(t) = v_1 \left(\frac{v_2}{v_1} \right)^\frac{t}{t0} $$

with $v(0)=v_1$ and $v(t0)=v_2$. The distance traveled is

$$ x(t) = \int_0^t \frac{1}{v(t)}\,{\rm d} t = \frac{t0 \left(1-\left(\frac{v_2}{v_1}\right)^{-\frac{t}{t0}}\right)}{v_1 \ln\left( \frac{v_2}{v_1} \right)} $$

But then you cannot specify the distance traveled as it is calculated with $ d_0 = x(t0) = \frac{t0 (v_2-v_1)}{\ln\left(\frac{v_2}{v_1}\right)} $


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