Annotate Instantaneous Velocity Problem

Recall General Wolgemuthian Method:

    1. Write the appropriate equation(s) - DO NOT substitute numbers of variables at the earliest possible time. I'm VERY serious about this. Let's discuss
    2. Isolate the variable you need to solve (unless it is already isolated)
    3. Substitute initial conditions into equations
    4. Show your answer


Average velocity vs instantaneous velocity.

Let's now please consider the following situation:

1) Let's say that you start at some position (we *usually* refer to that position as the origin) at time = 0.0

2) You travel 2.50 meters to the west in 1.00 seconds

3) You then travel another 2.5 meters to the west in the next 1.00 seconds

4) During the 3rd segment of your trip, you travel an additional 5.00 meters west in the next 1.00 seconds

5) Finally you travel 12.0 more meters west in the final 1.00 second of your trip.

Construct a table of that data


Does it look like this?


Time (sec) 
Displacement (m west)


Before you even start graphing that data, determine the likely shape of your graph...

Explain why you expect your graph to have that shape...


Notice that velocity is CHANGING over time, it also seems to be changing at a relatively constant sorta rate which indicates a smooth curve. Something like this perhaps?

Notice that I have included the original data points in light gray and an exponential - type "trend line" in red.

Now please calculate the average velocity between time 0.00 seconds and time 3.00 seconds

Notice that we begin by drawing a straight line to connect those to points on the curve.

We'll then work to find the velocity of that straight line. Since the slope of a displacement vs time graph is velocity, we'll use that to find the AVERAGE velocity between those two points.

Notice the difference between calculating AVERAGE velocity (connect two points on a curve) with INSTANTANEOUS velocity (draw a straight line TANGENT to ONE point on the curve).

Full Wolgemuthian ALWAYS:

Initial Conditions:

di = 0.0 m/s (you start timing at the origin)

df = 10.00 m west

ti = 0.0 sec (you start timing at some arbitrary origin)

tf = 3.00 seconds


v = (df-di) / (tf - di)


v is already on one side of the equal sign, no need to isolate


v = (10.00 m west - 0 meters west) / (3.00 sec - 0.0 sec)


3.33 m/s west (sig figs count *always*... so do units!)


Compare that with the velocity between 2.00 seconds and 4.00 seconds

Once again we begin by drawing a straight line to connect those two points on the curve. The slope of that line will give us average velocity between those two points.

There's two different errors/omissions on this graph, what are they?

Notice that you are actually calculating the slope between those points!!!

What's going on here? How can we have different velocities between different points?


The answer is we are calculating the AVERAGE velocity during those points.

Since the velocity itself is changing (increasing) it follows that the average velocity between two points on that curve would itself be changing.


With that in mind, how do you suppose we find the INSTANTANEOUS velocity of an object from a graph? hmmmmmmmmmmmmm -- talk it up!


Notice that the instantaneous velocity of an object can be obtained by find the SLOPE of the line at that instant point in time.... let's investigate

Let's say we want to find the instantaneous velocity of the object above right at 3.50 seconds.

We'd start by drawing a *tangent line* to the curve right at that time. In other words, we are going to try to precisely match the slope of that curve with the slope of a solid line, right at that point:

Now we find the slope of the tangent line. Since the point at time 3.50 seconds is also on the tangent line, the slope of the tangent line is also the value of the slope at 3.50 seconds. Since the slope of a velocity vs time graph is velocity, we have just found the instantaneous velocity at that point.!