One of the most extraordinary things about calculus is how it gives us a new way to connect concepts and one of the most beautiful is position, velocity and acceleration. We know we can start at a point, jog forward at some speed, and end up in a second spot. If we make the journey again, now running at a higher speed, we can travel farther in the same amount of time (change in distance, holding time constant) or arrive at our destination more quickly (change in time, holding distance constant). We can also alter our speed by speeding up or slowing down during the run itself, changing our acceleration.
These two changes, position influenced by velocity and velocity influenced by acceleration, can be modeled using derivatives. We start with an equation for position, in this case
f(x)=-(x-2)2+4.
This results in a lovely parabolic curve that starts at a height of zero at time zero and then goes up reaching a height of four at time two and landing back on the ground at time (height of zero) at time four. Try to describe the velocity in words. Remember that velocity can be negative.

We go up from time 0 to 2 and down from time 2 to 4. Our forward velocity is then positive from 0 to 2 and negative from 2 to 4. We calculate the velocity and graph it.

The velocity is the purple line. Desmos will graph derivatives for you: you can define your position with a function like F(x) then go to the next line and type
G(x) = d/dx F(x)
to see what it looks like (we will need the G(x) when we look at acceleration. Notice when the purple graph is positive (time 0 to 2) and negative (time 2 to 4) for the same intervals we said the object was traveling up and down. The very highest position is where the velocity crosses y=0. This is going to be super important later: at the top of a peak or bottom of a valley the first derivative will be zero because that is the point where you change direction. And what is the first derivative?
G(x)=-2(x-2)(1)=-2x+4.
Note that this is the equation of a straight line with a slope of -2 and a y-intercept of 4.
Now let’s look at the acceleration by taking the derivative of the velocity, G(x). The derivative of a derivative is called the second derivative of the initial function and it is also going to be very important (we can keep taking derivatives until we get zero, but most of our work as undergraduates will be with the first and second derivatives).

The acceleration is now green … and it’s a negative constant! This makes sense mathematically as
d/dx ( -2x+4) = -2.
The slope of the velocity is the acceleration. What does constant acceleration look like? You’ll probably find out in your physics one lab but if you can’t wait, find out if your school has a programable model car your classmates and you can borrow. Find a long flat quiet space and set the car loose at a very slow acceleration. See how long it takes before it is zooming at top speed!
So your teacher may ask you, when is the object speeding up and when it is slowing down? In this case, we can see from the graph that the object slows down as it gets higher until the peak, at which point it speeds up as it get closer to the ground. The point where the velocity is zero? That’s the critical point on our graph. Finding where velocity is zero is key to determining when a graph is slowing down or speeding up … but as you can guess, acceleration plays a part too. Next time we will look at what happens when acceleration isn’t constant!