Calc question with speed!?

Question

Please answer this question:
http://answers.yahoo.com/question/index;_ylt=AgpL0xEeObvJstlxDaVMtCrty6IX?qid=20070507172357AAtm4zj

While i appreciate explanations -> please actually do this ^^

Answer 1

Repeating the question so I don't have to refer back to the other window:

The position of a particle moving along a line is given by s(t)=2t^3+24t^2+90t+7 for t greater than or equal to 0. For what values of t is the speed of the particles increasing?

So, the velocity v(t) is given by v(t) = s'(t) = 6t^2 + 48t + 90 = 6(t^2 + 8t + 15) = 6(t+3)(t+5).
The speed is increasing when
(i) the velocity is non-negative and increasing, or
(ii) the velocity is non-positive and decreasing.

The acceleration is a(t) = v'(t) = 12t + 48 = 12(t+4).

The velocity is negative on (-5, -3) and positive on (-∞, -5) and on (-3, ∞). It is increasing on [-4, ∞) and decreasing on (-∞, -4].

So the speed is non-negative and increasing on [-3, &infin); and non-positive and decreasing on [-5, -4]. Hence the speed is increasing on these intervals.