v = v

x = x

v

Most school books will show "a" instead of "a

Unfortunately it is rarely mentioned that those simplified formulas are only applicable in special cases. The real formulas are even more simple and easier to remember than those taught in high school but they require a bit of higher math. You don't need to be a calculus whiz to get the idea behind the real formulas.

An object sitting at a fixed position has zero velocity. If you move it then it will have a certain velocity dependent on how fast it moves. In other words the

∂x(t) v(t) = ------- ∂tThe short notation is to put a dot on top of x(t) to mean the derivative. I don't have a dot in HTML therefore I use a dash:

∂x(t) v(t) = x'(t) = ------- ∂tThis simply a "funny notation" to say: draw the position (x) as a function of time, x(t), and measure the slope. The result is the velocity as a function of time. In other words

A change in velocity requires acceleration (or breaking=negative acceleration). Thus the slope of the velocity curve over time is acceleration over time:

∂v(t) ∂The reverse of a derivative is an integral. In graphical terms that is the area under a curve. For example a car driving with the constant speed of 2m/s (two meters per second):^{2}x(t) a(t) = v'(t) = ------- = x"(t) = -------- ∂t ∂t^{2}

v(t) = 2 m/s (blue line in the graph below).

The area under this blue line is a rectangle and it increases as you move to the right of the graph. This area corresponds to the position of our car. At t=2s the area is 2s*2m/s=4m, at t=3s the area is 2m/s*3s=6m.

x(t) = 2 t

If we draw 2*t into the graph then we get the red line. This is the position of the car over time.

This area under the velocity curve is called in mathematical terms an integral and written as:

x(t) = ∫v(t)∂t

The accumulated area over time under the velocity is the position. This is the reverse of "the change in position is velocity".

Change in velocity was acceleration and the reverse function would be "the area under the acceleration curve is velocity:

v(t) = ∫a(t)∂t

∂x(t) v(t) = x'(t) = ------- ∂t ∂v(t) ∂^{2}x(t) a(t) = v'(t) = ------- = x"(t) = -------- ∂t ∂t^{2}and the reverse functions: x(t) = ∫v(t)∂t + x_{0}= ∫∫a(t)∂t^{2}+ x_{0}v(t) = ∫a(t)∂t + v_{0}

∂f(t) function: f(t) = t^{2}, derivative: f'(t) = ------- = 2 t ∂t ∂f(t) function: f(t) = 2 t, derivative: f'(t) = ------- = 2 ∂t function: f(t) = constant, derivative: f'(t) = 0

In general the derivative of f(t) = t

The integral is the reverse operation of the derivative.

function: f(t) = t, integral: ∫f(t)∂t = ½ t^{2}function: f(t) = constant, integral: ∫f(t)∂t = constant * t

In general the integral of f(t) = t

The following diagram contains 3 graphs:

f

f

f

The graphs were chosen such that: ∫f

The integral of f

- acceleration is constant
- velocity increases linear from v
_{0}to v over the time period Δt

v = v

We get to that if we use

v(t) = ∫a(t)∂t + v

and set a(t) = a

The integral of a constant over time is the constant times the time: ∫a

x = x

This can be derived from this equation:

x(t) = ∫v(t)∂t + x

The assumption under which the high school equation is valid is: velovity increases linear v

The term ∫v(t)∂t would be the area marked in orange. That area is:

v

Replacing ∫v(t)∂t with ½ (v

Sorry, I am out of time. To be done later.

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