I have a background in calculus but don't really know anything about physics. Forgive me if this is a really basic question.
The equation for distance of an accelerating object with constant acceleration is:
$$d=ut +\frac{1}{2}at^2$$
which can also be expressed
$$d=\frac{\mathrm{d}x}{\mathrm{d}t}t+\frac{\mathrm{d^2}x}{\mathrm{d}t^2}\frac{t^2}{2}$$
(where x(t) is the position of the object at time t)
That's fine for a canonball or something like that, but what about a car accelerating from 0 to cruising speed? The acceleration is obviously not constant, but what about the change in acceleration? Is it constant? I suspect not. And then what about the change in the change of acceleration, etc. etc.? In other words, how does one know how many additional terms to add in the series?
$$d=\frac{\mathrm{d}x}{\mathrm{d}t}t+\frac{\mathrm{d^2}x}{\mathrm{d}t^2}\frac{t^2}{2}+\frac{\mathrm{d^3}x}{\mathrm{d}t^3}\frac{t^3}{3}+\frac{\mathrm{d^4}x}{\mathrm{d}t^4}\frac{t^4}{4}\cdot etc. \cdot ?$$
Best Answer
There are three cases here:
The acceleration is a function of time $a(t)$. Then the velocity is $$v(t)=v_c+\int a(t)\,{\rm d}t \tag{1}$$ and the position as a function of time $$x(t)= x_c + \int v(t)\,{\rm d}t \tag{2}$$ The distance is calculated from $x(t)$.
The acceleration is function of position $a(x)$. Then the velocity as a function of position is $$ \frac{1}{2}v(x)^2 = w_c + \int a(x)\,{\rm d}x \tag{3}$$ and the time as a function of position $$ t(x) = t_c + \int \frac{1}{v(x)}\,{\rm d}x \tag{4}$$ which needs to be back-solved for $x(t)$.
Where $x_c$, $v_c$, $t_c$ and $w_c$ are integration constants of appropriate units
Example 1
$ a(t) = -100 \sin(10 t)$, with $x(0)=0$ and $v(0)=10$ $$ v(t) = \int -100\sin(10 t)\,{\rm d}t = 10\,\cos(10 t) $$ $$ x(t) = \int 10\cos(10 t)\,{\rm d}t= \sin(10 t)$$
Example 2
$ a(x) = -100 x$, with $x(0)=0$ and $v(0)=10$ $$ \frac{1}{2}v(x)^2 = \int -100 x {\rm d}x = 50 (1-x^2) $$ $$ v(x) = 10 \sqrt{\left(1-x^2\right)} $$ $$ t(x) = \int \frac{1}{10 \sqrt{\left(1-x^2\right)}}\,{\rm d}x = \frac{\sin^{-1}(x)}{10} $$ $$ x(t) = \sin(10 t) $$
Example 3
$ a(v) = 100 - 5 v $, with $x(0)=0$ and $v(0)=10$ $$t(v) = \int \frac{1}{100 - 5 v}\,{\rm d}v = -\frac{1}{5}\ln{ \left( \frac{20-v}{10} \right) } $$ $$x(v) = \int \frac{v}{100 - 5 v}\,{\rm d}v = 2-\frac{v}{5}-4 \ln{\left(\frac{20-v}{10}\right)} $$ with solution $v(t) = 20-10 \hat{e}^{-5 t}$ and $x(v(t)) = 2 \hat{e}^{-5 t}+20 t-2 $