# How to determine an exponential acceleration curve?

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.

• You are just not using suitable integration constants. – 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: $$x(t) = x_1 + (v_1-\tfrac{A}{k}) t + \frac{A}{\hat{k}^2}\exp(\hat{k}t)\\$$

• Something must me goofy because the last equation doesn't have the right units... – kηives Jun 6 '12 at 16:13
• true, goes right from the $\exp$ assumption, let's introduce a factor – IljaBek Jun 6 '12 at 16:17
• 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. – Emilio Pisanty Jun 6 '12 at 18:09
• The numerical solution can also be avoided using tables for the product log function: wolframalpha.com/input/?i=%28e%5Ex-1%29%2Fx+inverse – Emilio Pisanty Jun 6 '12 at 18:13
• @EmilioPisanty: correcting, thx. good tip, but numerical solution stays numerical ;) – 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 $d_0 = x(t0) = \frac{t0 (v_2-v_1)}{\ln\left(\frac{v_2}{v_1}\right)}$