Wednesday, November 19, 2014

Suppose the position of an object at time t is noted by the function s(t) = -4.9t^2 + 300.  The object is dropped from the top of a building. What is the average velocity of the object in the first three seconds after it is dropped?

The velocity, v(t), is found by taking the derivative, which is the rate of change of the object.

v(t) = -9.8t

Now we can used the mean value theorem, which is [f(b)- f(b)]/(b - a) for some interval a to b in which the function is differentiable. The interval is 0 to 3, so a = 0 and b = 3

f(3) = 255.9
f(0) = 300

(255.9 - 300)/(3 - 0) = -14.7 which is the average velocity. How can velocity be negative, one might think? The object is going downward, which makes it negative. Speed can only be positive but velocity can be both positive or negative depending on the direction.

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