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?
$$d=\frac{\mathrm{d}x}{\mathrm{d}t}t+\frac{\mathrm{d^2}x}{\mathrm{d}t^2}\frac{t^2}{2}+\frac{\mathrm{d^3}x}{\mathrm{d}t^3}\frac{t^3}{3}+\frac{\mathrm{d^4}x}{\mathrm{d}t^4}\frac{t^4}{4}\cdot etc. \cdot ?$$
Answer
Technically, the equation
$$d = \frac{\mathrm{d}x}{\mathrm{d}t}t + \frac{\mathrm{d}^2x}{\mathrm{d}t^2}\frac{t^2}{2}$$
is not right. Instead, for constant acceleration, you need
$$d = \left(\left.\frac{\mathrm{d}x}{\mathrm{d}t}\right|_0\right) t + \left(\left.\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\right|_0\right) \frac{t^2}{2}$$
In other words, a quantity like $\mathrm{d}x/\mathrm{d}t$ changes in time, but you want to use the initial velocity only. I think this is what you probably intended to begin with, though.
If you wanted to solve the problem purely kinematically, then you could try to expand the position in a Taylor series as you wrote in your answer. However, this only works if the function is equal to its Taylor series. For simple functions like exponentials and trig functions this is true, but for a person driving a car it is not. If a function equals its Taylor series everywhere, then if you observe its position over any finite interval of time, no matter how short, you can completely determine what the car will do in the future. This is not realistic.
Instead, you will want some way of determining either the velocity or the acceleration as a function of time or position. In physics, it is common to be able to determine the acceleration as a function of position. The reason is that acceleration comes from the equation $$F=ma$$ so that if you can determine the forces present, you know the acceleration, and higher-order derivatives are not necessary.
If you know the velocity as a function of time, you can simply integrate it to find the displacement. $$d(t) = \int_{t_0}^t v(t') \mathrm{d}t'$$
If you know the acceleration as a function of time, you can integrate that too, although this situation is less common.
$$d(t) = v_0(t - t_0) + t\int_{t_0}^t a(t')\mathrm{d}t' - \int_{t_0}^t t'a(t')\mathrm{d}t'$$
I found this expression by looking for something whose derivative with respect to time was the velocity
$$v(t) = v_0 + \int_{t_0}^t a(t')\mathrm{d}t'$$
If you know the velocity as a function of position, you have the differential equation
$$\frac{\mathrm{d}x}{\mathrm{d}t} = v(x)$$
which you can solve by separation of variables.
If you know the acceleration as a function of position, you have the differential equation
$$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = a(x)$$
which is not always easy to solve. In more realistic scenarios, the acceleration will depend not only on the object's own position, but also on the positions of the things it's interacting with. This gives coupled differential equations, which can be simplified in a special cases, but frequently can only be solved numerically.
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