Integral as Net Change

Problem 1:
A particle moves along the x-axis so that its acceleration at any time t is given by a(t) = 6t - 18. At time t = 0 the velocity of the particle is v(0) = 24, and at time t = 1 its position is x(1) = 20.
    (a) Write an expression for the velocity v(t) of the particle at any time t.
    (b) For what values of t is the particle at rest?
    (c) Write an expression for the position x(t) of the particle at any time t.
    (d) Find the total distance traveled by the particle from t = 1 to t = 3.

Answers:

v(t) = ò a(t) dt = ò (6t - 18) dt = 3t² - 18t + C

24 = 3(0)² - 18(0) + C

24 = C

v(t) = 3t² - 18t + 24

(b) The particle is at rest when v(t) = 0.
       3t² - 18t + 24 = 0
       t² - 6t + 8 = 0
       (t - 4)(t - 2) = 0
       t = 4, 2
(c) x(t) = ò v(t) dt = ò (3t² - 18t + 24) dt = t³ - 9t² + 24t + C
      20 = 1³ - 9(1)² + 24(1) + C
      20 = 1 - 9 + 24 + C
      20 = 16 + C
       4 = C
      so x(t) = t³ - 9t² + 24t + 4

(d) To calculate the distance traveled by the particle, the turning points must be
considered. The turning points are x = 2, 4. Check sign changes.

            +          -
        |---------|---------|------------   v(t)
       1           2          3

        Since the velocity changes from positive to negative at t = 2, the particle stops and
        reverses its direction, so the distance from t = 1 to t = 2 must be calculated, and the
        distance from t = 2 to t = 3 must be calculated, and the two distances must be added
        together.

        Method 1:                                                      Method 2:
        x(1) = 20                                      2                                  3
        x(2) = 24                                     ò (3t² - 18t + 24) dt + ò (3t² - 18t + 24) dt
        x(3) = 22                                    1                                 2

        | x(2) - x(1)| = 4                           The area under the velocity curve is rate x time
        | x(3) - x(2)| = 2                            which gives distance. Calculating the integrals
         Total distance = 6                         gives us the area under the curve. Distance is
                                                             being accumulated. Distance is positive, so use the
                                                             absolute value of any negative integrals.
                                                                   2                              3
                                             t³ - 9t² + 24t |    +   | t³ - 9t² + 24t | |
                                                                  1                              2

= (8 - 36 + 48) - (1 - 9 + 24) + |(27 - 81 + 72) - (8 - 36 + 48)|

= 20 - 16 + |18 - 20|

= 6

Problems