Rate of Change in Natural & Social Sciences

Question

A particle moves along a straight line and its position at time t is given by s(t)=t3-9t2+15t where s is measured in feet and t in seconds. Consider t in the interval [0,infinity).

1. when is the particle moving in the positive direction? (enter answer in interval)
2. what is the total distance the particle travels between 0 and 9?
3. when is the particle slowing down? (enter in interval)

Answer 1

s(t) = t³ - 9t² + 15t

s'(t) = 3t² - 18t + 15 = 3(t² - 6t + 5) = 3(t - 5)(t - 1)

The particle moves in the positive direction when s'(t) > 0.

s'(t) = 0 when t = 1 or 5.

When 0 < t < 1, s'(t) > 0. So, the particle is moving to the right.

when 1 < t < 5, s'(t) < 0. So, the particle is moving to the left

when t > 5, s'(t) > 0. So, the particle is moving to the right.

So, the particle moves in the positive direction when 0<t<1 and when t>5.

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Total distance from t = 0 to t = 9: [s(1) - s(0)] + [s(1) - s(5)] + [s(9) - s(5)]

= [7 - 0] + [7 - (-25)] + [135 - (-25)] = 199 feet

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The particle slows down when s'(t) and s"(t) have opposite signs.

s"(t) = 6t - 18

s"(t) = 0 when t = 3

When 0 < t < 3, s"(t) < 0 and when t > 3, s"(t) > 0.

So, the particle slows down when 0 < t < 1 and when 3 < t < 5.

Answer 2

The easiest way to do this problem is to graph s(t) and compute v(t)=s'(t)=

3t²-18t+15 and graph it as well.

The velocity is positive on [0,1]and on [5,∞), i.e. the particle is moving in the positive direction.

To get the distance traveled you need to integrate v(t) in 3 parts and add the absolute values of those 3 parts:

[0,1] [1,5] and [5,9] because the distance traveled between 5 and 9 will be algebraically negative.

The particle is slowing on [0,3]

You can, of course, do this algebraically, but it is just easier to see graphically. And if you need to give an algebraic answer, the graph will help make sure your answer is correct.