Kinematics

Calculus was "created" to solve problems, although it now seems to create problems for students. Some of these problems involved physics. As more of a physics guy, I like to believe that physics was the major motivator, at least for Newton. Many problems in physics could not be solved without the ideas of calculus. The most basic of which were Kinematic problems, that is problems with quantitatively describing motion.

### Position, Velocity and Acceleration

Lets start with the most basic situation. Imagine a car traveling in a straight line at a constant speed of 2 m/s. That means that for every 1 second the car's position will change by 2 m. If we graph that we get something like:

Notice that the slope of the line is equal to $slope=\frac{7-1}{3-0}= 2 \: \: m/s$. Which is exactly the speed of the object. We are once again talking about slope which should bring to mind the idea of derivative…

If we define a function $s(t)$ that describes the position of an object as a function of time. Then the velocity of the object whose position is described by $s(t)$ will be the slope or derivative of $s(t)$.

(1)
\begin{align} v(t)=s'(t)=\frac{ds}{dt} \end{align}

The derivative is over-kill when the position function is linear. However if we take a more complicated example such as an object dropped (starting at rest) and falling only under the influence of gravity

(2)
$$s(t)=-4.9t^2$$

Notice that at $t=0$ the position of the ball is zero. Which is fine, but it can make some folk uncomfortable when the position becomes negative. To help those folks, lets assume the object starts 10m above the ground. The modified equation is:

(3)
$$s(t)=-4.9t^2+10$$

Looking at the graph we can clearly see that the slope of the line increases as the object falls (note that I have adjusted the axis to exaggerate the shape).

The derivative of this new function is $v(t)=-9.8t$. Note that the adjustment of the initial height does not affect the derivative function! If we graph the velocity vs. time function we will again get a straight line with a constant slope. The slope of this line is -9.8 with units that look like $\frac{m/s}{s}=m/s^2$. Which happen to be the unit of acceleration. It turns out that the slope of a velocity vs. time graph is the acceleration.

So…

(4)
\begin{align} a(t)=v'(t)=\frac{dv}{dt} \end{align}

Since $v(t)$ is the derivative of $s(t)$ then we can also define acceleration as the second derivative of position.

(5)
\begin{align} a(t)=s''(t)=\frac{d^2s}{dt^2} \end{align}

This also means that the acceleration (in this case) is a constant function given by:

(6)
$$a(t)=-9.8$$

### Going the Other Way…

The concepts of position, velocity and acceleration are related through the idea of a derivative. It seems reasonable to assume that if you can go from velocity to acceleration that you should be able to go from acceleration to velocity. Lets continue with the example above. The graph of acceleration vs. time would look something like:

From this graph we should be able to discover something about the velocity… Well it turns out that the "area under the curve" tells us about the velocity. You can partially justify this by seeing that the longer the object was to fall the larger the area under the curve since the function would extend further and further. As the object falls more and more its velocity will also increase. That's not a proof, but at least the two things I'm saying are related both increase…

In the graph above we can calculate the area under the curve easily as it is a rectangle. Note: The phrase "area under the curve" refers to the area between the function and the horizontal axis." The width of the rectangle is approximately 1.43 and the height is -9.8. Giving a total area of -14.0. Which is the same value as the velocity in the graph above at $t=1.43$.

We can continue the argument with a velocity vs. time graph. It seems reasonable that a velocity vs. time graph would tell us something about the position. It turns out that the area under the curve of a velocity vs. time graph is the position… Looking again at the velocity vs. time graph:

This time the area is a triangle with the $area=\frac{1}{2} (1.43)(-14)=-10$. Which is the same distance that the object fell! The negative simply indicates direction, i.e. towards negative. Call me a nerd, but that's pretty cool.

### Area under the Curve

This is a huge topic. One that is often even more useful than the derivative. It is the idea of the "anti-derivative" that is an operational that "undoes" a derivative. This is also very commonly call an integral or integration.

In summary the relationships be represented by the following diagram: