Calculus I (AB) and II (BC): That Pesky Particle Assume that a particle (all hopped up on caffeine) is moving along the y-axis, and its velocity (in ft/sec) at any time t (in seconds) is given by .
    (a)    If its position at time t = 0 is two units below the origin, find the position equation x(t).
    (b)    What is the particle's average accleration on the interval t = [2,5] ? Include units.
    (c)    What is the particle's average position on the same interval?
    (d)    How many times does the particle change direction on the t interval [0,10]?
    (e)    When, during the t interval [0,10], is the particle moving upward?
    (f)    What is the total distance traveled by the particle on [0,10]?

Difficulty Rating: 

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Solution:

    (a) The antiderivative of v is position, so Now, you know that x(0) = 2 from the given info, so

    (b) The average acceleration is given by the slope of the secant line to velocity on t = [2,5].

(You could also have found the derivative of velocity, which is acceleration, and used the average value of a function formula.)

    (c) To calculate average position, find the average value of the position equation.

    (d) The particle may change direction when it stops (when velocity is 0). However, you must verify that the direction actually changes, which you can do by examining the sign of the velocity (in this problem, positive velocity indicates upward movement, and negative velocity, downward movement). All of this can be accomplished with a wiggle graph of velocity. Begin by finding critical points for velocity:

Uh oh. This equation does not have any solutions, and therefore, the velocity never equals 0, and the particle never changes direction! Thus, the particle only goes in one direction the entire time. Plug any point on t = [0,10] into the derivative to determine the sign (and hence the direction of the particle on that interval). The particle always moves upward.

    (e) We just answered that question, dadgum it.

    (f) Since the particle doesn't change direction, the total distance traveled is
x(10) - x(0) = 2948  (2) = 2950 feet.