Calculus Applets |
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Let's take a look at average velocity. If you recall from earlier mathematics studies, average velocity is just net distance traveled divided by time. For example, if an object is tossed into the air we might find the following data for the height in feet, y, of the object as a function of the time in seconds, t, where t = 0 is when the object is released upward.
| t (sec) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| y (feet) | 6 | 90 | 142 | 162 | 150 | 106 | 30 |
This table says that the object was 6 feet above the ground when released,
90 feet after 1 second, etc. How fast was it traveling? We can divide the net
distanced traveled by the time to compute the velocity. For example, over the
first second the average velocity is
. Over the last second, the
average velocity is
. Note that velocity can be
negative to indicate downward motion (positive velocity is upward motion in
this problem). Speed is just the magnitude of the velocity (i.e.,
the absolute value, in our one-dimensional example). We can also compute the
average velocity over the entire interval as
.
Now, what if we wanted to know the velocity at a specific time, say at t = 1? Clearly there is some velocity, as the object is moving, but to
compute velocity we need a ratio of distance traveled to time spent
traveling. We could find the average velocity over the interval from 1 to 2
seconds, which is just
. But this isn't quite right, as
the object does seem to be speeding up. If we had another measurement of
height at a time closer to 1 than 2 seconds, we could get a better estimate
of the instantaneous velocity at t = 1. The following applet
illustrates this.
Try the following:
This work by Thomas S. Downey is licensed under a Creative Commons Attribution 3.0 License.
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