When an object moves, its position changes. If the object moves along a line (whether the line is horizontal, vertical or oblique), we can attempt to describe the motion of the object by giving a function x(t) which reports the displacement x of the object in convenient units from some origin at time t. The function might be described by a formula, by a table of values, or by a graph.
For example, let's think about the vertical motion of a small object
(a cantaloupe, say) that is thrown into the air at t = 0
seconds. The
general form of the motion of the object is obvious: it goes up,
slows down, reverses direction, falls down faster and faster and
finally, Splat! This gives a general description of the
velocity of the melon, but can we be more precise? For instance,
assume that we can measure the height of the melon at any time
t. Think of the height x
above the ground as a function of time:
| t(secs) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| x(feet) | 6 | 90 | 142 | 162 | 150 | 106 | 30 |
Splat! comes sometime between 6 and 7 seconds. The numbers show the behavior noted above - during the first second, the melon travels 90 - 6 = 84 feet, but in the second second it travels only 142 - 90 = 52 feet. Since speed = distance/time, we can say that the average speed over the first second is 84 feet/second and the average speed over the time interval 1 < t < 2 is 52 feet/second. In general, the average velocity of a moving object over the time interval a < t < b is the net change in position x(b) - x(a) divided by the change in the time b - a. The sign of the (average) velocity has significance!

Do exercise
2.1.
The average velocity gives a rough idea of the behavior of the melon, but average velocity over an interval does not solve the problem of determining the velocity of the melon exactly at t = 1, say. To get closer to an answer to that question, we have to look at what happens near t = 1 in more detail.
To do this, suppose that the melon incident was videotaped, and we can run the film against a precise stopwatch. We could make some more measurements of the height of the melon at times near t = 1 sec:
If we compute the average velocity of the melon over the time interval [0.9, 1], we find it to be
(90 - 83.04)/(1 - 0.9) = 69.6 ft/s
Over the interval [1, 1.1] we get:

To be even more precise, we could make measurements over shorter time
intervals around 1 sec:

You should do exercise
2.2 before
proceeding...
What we just did with velocities is the general idea of computing derivatives - we wanted the instantaneous velocity at t = 1, but we could only compute average velocities. To arrive at the instantaneous velocity at a given time, we took average velocities over shorter and shorter time intervals containing that time.
There is no reason to stick to t = 1 - we just happened to have lots of data about the flying melon around that time. In mathematics, one usually begins with a function that is defined for a whole range of values, so it can be sampled as much as you want. For instance, all the data given above was in fact calculated from the formula: x = 6 + 100t - 16t² (you had already guessed that, right?), so we could compute instantaneous velocities at other times in a similar way.

Exercise
2.3 leads you
through a more accurate representation of the instantaneous
velocity using Maple.
For more on treating special relativity, see the online notes of Dr. Park.

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