Splat! comes sometime between 2.5 and 3 seconds. The numbers don't quite show the behavior noted above because the canteloupe travels upward for less than half a second. But the numbers reveal the acceleration of the canteloupe toward the ground: During the first second, the net displacement of the canteloupe is 84 - 95 = -11 feet (note that we count upward displacement as positive and we place the origin at ground level), but in the second second its displacement is 41 - 84 = -43 feet. Since speed = distance/time, we can say that the average speed over the first second is 11 feet/second and the average speed over the time interval 1 < t < 2 is 43 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. Therefore the average velocity of the canteloupe between time t = 0 sec and time t = 1 sec is (84 - 95)/(1 - 0) = -11/1 = -11 feet per second. Similarly, the average velocity of the canteloupe between time t = 1 sec and time t = 2 sec is -43 feet per second. The sign of the (average) velocity has significance! In this case it indicates that the canteloupe is moving downward.
Do exercise 2.1.
The average velocity gives a rough idea of the behavior of the canteloupe, but average velocity over an interval does not solve the problem of determining the velocity of the canteloupe 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 canteloupe incident was videotaped, and we can run the film against a precise stopwatch. We could make some more measurements of the height of the canteloupe at times near t = 1 sec:
If we compute the average velocity of the canteloupe over the time interval [0.9, 1], we find it to be
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 canteloupe 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 = 95 + 5t - 16t2 (you had already guessed that, right?), so we could compute instantaneous velocities at other times in a similar way.
A summary of the difference between average and instantaneous quantities for velocity can be found here.
To test your general understanding of the relationship between velocity and acceleration, here is an excellent Java applet for you to work with.
Exercise 2.3 leads you through a more accurate representation of the instantaneous velocity using Maple. Also check out Exercise 2.4
For more on treating special relativity, see the online notes of Dr. Park.
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