Calculus AB and Calculus BC
CHAPTER 4 Applications of Differential Calculus
J. MOTION ALONG A LINE
Velocity
Acceleration
Speed
If a particle moves along a line according to the law s = f (t), where s represents the position of the particle P on the line at time t, then the velocity v of P at time t is given by and its acceleration a by . The speed of the particle is |v|, the magnitude of v. If the line of motion is directed positively to the right, then the motion of the particle P is subject to the following: At any instant,
(1) if v > 0, then P is moving to the right and its distance s is increasing; if v < 0, then P is moving to the left and its distance s is decreasing;
(2) if a > 0, then v is increasing; if a < 0, then v is decreasing;
(3) if a and v are both positive or both negative, then (1) and (2) imply that the speed of P is increasing or that P is accelerating; if a and v have opposite signs, then the speed of P is decreasing or P is decelerating;
(4) if s is a continuous function of t, then P reverses direction whenever v is zero and a is different from zero; note that zero velocity does not necessarily imply a reversal in direction.
EXAMPLE 26
A particle moves along a line such that its position s = 2t^{3} − 9t^{2} + 12t − 4, for t 0.
(a) Find all t for which the distance s is increasing.
(b) Find all t for which the velocity is increasing.
(c) Find all t for which the speed of the particle is increasing.
(d) Find the speed when
(e) Find the total distance traveled between t = 0 and t = 4.
SOLUTION:
and
Velocity v = 0 at t = 1 and t = 2, and:
Acceleration a = 0 at and:
These signs of v and a immediately yield the answers, as follows:
(a) s increases when t < 1 or t > 2.
(b) v increases when
(c) The speed |v| is increasing when v and a are both positive, that is, for t > 2, and when v and a are both negative, that is, for
(d) The speed when equals
FIGURE N4–15
(e) P’s motion can be indicated as shown in Figure N4–15. P moves to the right if t < 1, reverses its direction at t = 1, moves to the left when 1 < t < 2, reverses again at t = 2, and continues to the right for all t > 2. The position of P at certain times t are shown in the following table:
t: |
0 |
1 |
2 |
4 |
s: |
−4 |
1 |
0 |
28 |
Thus P travels a total of 34 units between times t = 0 and t = 4.
EXAMPLE 27
Answer the questions of Example 26 if the law of motion is
s = t^{4} − 4t^{3}.
SOLUTION: Since v = 4t^{3} − 12t^{2} = 4t^{2} (t − 3) and a = 12t^{2} − 24t = 12t(t − 2), the signs of v and a are as follows:
Thus
(a) s increases if t > 3.
(b) v increases if t < 0 or t > 2.
(c) Since v and a have the same sign if 0 < t < 2 or if t > 3, the speed increases on these intervals.
(d) The speed when
FIGURE N4–16
(e) The motion is shown in Figure N4–16. The particle moves to the left if t < 3 and to the right if t > 3, stopping instantaneously when t = 0 and t = 3, but reversing direction only when t = 3. Thus:
t: |
0 |
3 |
4 |
s: |
0 |
−27 |
0 |
The particle travels a total of 54 units between t = 0 and t = 4.
(Compare with Example 13, where the function f (x) = x^{4} − 4x^{3} is investigated for maximum and minimum values; also see the accompanying Figure N4–5.)
BC ONLY