Position to Velocity

Tyler Seacrest

4 Position to Velocity

The idea of position of an object versus the velocity of an object encompasses all the big ideas of calculus. So that’s where we’ll start!

Heading to a lake

Suppose you’re given a graph of your distance from home during a trip to the lake. It might look something like this:

This graph might represent you walking to a lake two miles away, hanging out for half an hour, then walking home. The first part of the graph that slants upwards represents your walk to the lake, since your distance from home is increasing (higher on the graph). The second part of the graph represents you hanging out at the lake. It’s flat since your distance from home is not changing. Finally, the part of the graph that slants down represents you walking home. Your distance to home is decreasing, so the line goes down on the graph.

Now here is the question: what is your velocity during this journey?

Velocity is a measure of speed, and essentially boils down to this equation:

$Velocity = \frac{change in distance}{change in time} .$

While you’re walking to the lake, you’re traveling at a rate of 2 miles every half hour (your change in distance is two, during the half hour change in time). Therefore your velocity is $\frac{2}{\frac{1}{2}}$ . We can simplify this fraction by multiplying top and bottom by $2$ , and we see

$Velocity to the lake = \frac{2}{\frac{1}{2}} \cdot \frac{2}{2} = \frac{4}{1} = 4.$

So you were walking at 4 miles per hour to the lake. This velocity doesn’t change as you walk to the lake, and so we call this a constant velocity. Graphically, we represent constant velocity with a horizontal line:

While at the lake, your position is not really changing ’cause you’re just hanging out. So your velocity is zero. To relate this back to the formula, your change of distance is zero while your change in time is $1 / 2$ . So by the formula we have

$Velocity at the lake = \frac{0}{\frac{1}{2}} = 0,$

but we didn’t really need to do the formula since we knew we weren’t going anywhere. Graphically, a velocity of zero looks like this:

Finally, while coming back home, again we have a change in distance of $2$ miles over a half an hour. So you might think the velocity is $4$ again, but it is actually very natural to call this a negative velocity, since the distance is going down. So we say that the change in distance is actually $- 2$ , and therefore:

$Velocity returning home = \frac{- 2}{\frac{1}{2}} \cdot \frac{2}{2} = \frac{- 4}{1} = - 4,$

Graphically:

Okay, now let’s look at the position and velocity graphs together. I’ll color each segment to emphasize how the different parts correspond.

Slope

Another way to think of velocity is that it is the same as the slope of a line. Recall that the slope of a line is a measure of how steep the line is, and the formula follows the phrase “rise over run”. Let’s look again at the position and velocity graphs from the last subsection:

What is the slope of the red line? Well, rise over run would be $\frac{2}{0.5}$ , which is $4$ . That’s the same as the velocity graph! Same thing for the green line: it has a slope of zero, and the velocity graph is at zero. Finally, the slope of the blue line is $\frac{- 2}{0.5}$ which is $- 4$ , and that is what we have for the velocity graph.

So “slope” and “velocity” are the same thing. But there is another name for this concept that we will use a lot: derivative. Derivative, slope, and velocity all mean the same thing.

Other Examples of Derivatives

Let’s see some other examples. Note for each of these, the position graphs is always piecewise linear, or made up of line segments. This makes it easier to find the velocity, or slope.

Position to Velocity

Find the velocity graph (i.e. the derivative) corresponding to the following position graph.

To solve this problem, we need to find the velocity, or slope, of each of the lines in the graph. The first line has a change of distance of $60$ , and a change of time of $5$ seconds, so the velocity is $60 / 5 = 12$ . Next, the we have a negative change of distance of $- 40$ since the graph goes from $60$ to $20$ . This also occurs over $5$ seconds, so the velocity is $- 40 / 5 = - 8$ . Finally, we gain a distance of $20$ in the final line over five seconds, so the velocity is $20 / 5 = 4$ .

If we graph these velocities, we have

Integrals

We can also go in the reverse direction: take a velocity graph, and create a position graph. This is called integration or taking an integral. This can be tricky but we can do it at this point if the function is what is called a step function, which is basically a function consisting of a bunch of flat parts.

Integration

Given the following velocity graph, create a position graph.

Given these velocities, we want to graph where the person travels, and the person can start wherever we want. For convenience we will start the person at location zero. If we focus on the first section, we see the person is traveling at 5 miles per hour for three hours. This corresponds to the person traveling $15$ miles total during the first three hours. It would look something like this:

Notice how the position goes from $0$ to $15$ . From $3$ to $6$ , the person moves at $- 10$ miles per hour for $3$ hours: this would be a total of $- 30$ miles traveled. Since the person is already at position $15$ , they’ll end up at position $15 + (- 30) = - 15$ .

From $6$ to $9$ , the person moves at $- 5$ miles per hour for $3$ hours, which is another $- 15$ miles covered. Starting from position $- 15$ and adding another $- 15$ , the person will end up at $- 30$ .

Finally, here are colored versions of the velocity and position maps.

Going from velocity back to position is called an integral. Here is another example.

Integration

Given the following velocity graph, create the position graph.

To find out the position, note that we use multiplication. For example, in the first 3 hours, they move at 4 miles per hour, so we multiply: $3 \cdot 4 = 12$ miles. In the next $4$ hour stretch, we’re at $- 5$ mph, so we multiply $4 \cdot (- 5) = - 20$ . The last bit is two hours long at 12 mph, so we multiply $2 \cdot 12 = 24$ . Note that we ALSO multiply when we find area, so we can think of these calculations (velocity to position) as area calculations:

So we basically make jumps of $12$ , $- 20$ , and $24$ as shown:

Increasing and Decreasing

The following graphs are not made up of straight lines — but we can still tell if the derivative is positive or negative. A positive derivative means a quantity is increasing — and graphically that is represented by a graph the climbs as you go from left to right. A negative derivative means a quantity is getting smaller — graphically going downward from left to right.

Which of the following have positive derivatives? Which have negative derivatives?

(a)