Calculus AB Problem

1. Feb 17, 2015

icecubebeast

1. The problem statement, all variables and given/known data
The position of a particle moving along a line is given by s(t) = 2t^3 -24t^2 + 90t + 7 for t ≥ 0. For what values of t is the speed of the particle increasing?
(a) 3 < t < 4 only
(b) t > 4 only
(c) t > 5 only
(d) 0 < t < 3 and t > 5
(e) 3 < t < 4 and t > 5
2. Relevant equations
d/dx [k*x^n] = kn*x(n-1) power rule
d/dx [f(x) +- g(x)] = f'(x) +- g'(x)
3. The attempt at a solution
s(t) = 2t^3 -24t^2 + 90t + 7
find the derivative:
s'(t) = 6t^2 - 48t + 90
find the second derivative:
s''(t) = 12t - 48

Since the problem is asking for acceleration "For what values of t is the speed of the particle increasing?", we find the point of inflection and find the intervals.
s''(t) = 0
12t - 48 = 0
12(t - 4) = 0
t=4
Intervals: (0,4] and [4,infinity)
We plug in a number for each interval:
s''(1) = 12(1) -48
=12 - 48
=-36

s''(5) = 12(5) - 48
=60 - 48
=12

The values of t>4 are when the speed is increasing.

The problem is that the solution to this problem is (e) 3 < t < 4 and t > 5 and I don't know why?

2. Feb 17, 2015

DEvens

Hmm... Do you suppose the author of the question is thinking "magnitude of speed" rather than just speed?

A positive speed getting smaller is not speed getting bigger. This is certain. But is a negative speed getting more negative, is the speed increasing?

3. Feb 17, 2015

icecubebeast

4. Feb 17, 2015

DEvens

Oh, no. My physics teachers were perfect in every way.

5. Feb 17, 2015

icecubebeast

6. Feb 17, 2015

Ray Vickson

Standard nomenclature is that speed = absolute value of velocity. Your $s'(t)$ is velocity $v(t)$, not speed. If you plot the speed $|s'(t)|$ you will see how the book's answer arises.

If you want to do it without plotting, you need to be careful about when speed is increasing; this occurs if either (a) $v(t) < 0$ is decreasing, or (b) $v(t) > 0$ is increasing. Do you see why?

7. Feb 17, 2015

icecubebeast

So speed is different in that increasing speed depends only on the magnitude of the increase of speed, and not dependent on direction?

8. Feb 17, 2015

Ray Vickson

No: increasing speed means increasing magnitude of velocity. If velocity is < 0 and is becoming < 0 at an even greater rate, speed increases; if velocity is > 0 and is becoming > 0 at an even greater rate, speed increases. Just think about velocity being > 0 if we are driving East and being < 0 if we are driving West, and think about what increasing speed means in both cases.