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Note that in this problem, every one of these questions can be identified as asking about either position, velocity, or acceleration.

(a), (b), (c), and (d) are asking questions about velocity.

(e) is asking about the *change* in position but not the *rate* of change in position. Therefore this question is more relevant to position, not velocity.

Ignore parts (f) and (h).

(g) and (i) are about acceleration.

I will show you the solution to problem 1. You should solve problem 2. --- #### Problem 1: (a) ${tex`v(t)=3t^2-24t+36`} --- (b) ${tex`v(3)=3(3)^2-24(3)+36 = 27-72+36 = -9`} --- (c) ${tex.block`v(t) = 0 = 3t^2-24t+36 \\ \Rightarrow 0 = t^2-8t+12`} So we solve ${tex`0 = (t-6)(t-2)`} which has solutions 6 and 2. Therefore the particle is at rest when the time is 2 seconds and then when time is 6 seconds. --- (d) The particle is moving in the positive direction when ${tex`v(t)>0`} so we solve ${tex`{3t^2-24t+36 > 0}`}. To analyze this we use the solution above to find the intervals ${tex`(-\infty,2)`} and ${tex`(2,6)`} and ${tex`(6,\infty)`} but we don't know which of these are "positive intervals" (i.e. intervals on which ${tex`v(t)`} is positive). To determine that, we test each interval at some point inside. If we test ${tex`(-\infty,2)`} we may choose the point ${tex`t=0`}. There we find ${tex`v(0) = 36`} so this is a "positive interval". If we test ${tex`(2,6)`} at ${tex`t=3`} we find ${tex`v(3)=-9`} (In fact this is just the answer to part (b).) Therefore this is a "negative interval". If we test ${tex`(6,\infty)`} at ${tex`t=7`} we find ${`v(7)=5`} so this is again a positive interval. Initially it seems like the velocity is positive for time in the set ${tex`{(-\infty,2)\cup(6,\infty)}`}. However, since we restrict to ${tex`t\ge 0`} then we should say that the particle is moving in the positive direction on the set ${tex`[0,2)\cup (6,\infty)`}. --- (e) ${tex.block`s(8) = 8^3-12(8)^2+36(8) = 32`} --- (f) Skip. --- (g) ${tex`a(t) = 6t-24`} so ${tex`a(3)=18-24=-6`}. --- (h) Skip. --- (i) A function is speeding up in two cases: Its velocity is positive and velocity is increasing, or its velocity is negative and is decreasing. (Both of these mean "speeding up" because a very negative velocity corresponds to a fast motion in the negative direction.) We already know (from part (d)) that the velocity is positive on the set ${tex`{[0,2)\cup(6,\infty)}`}. We need to also find where it is increasing, which is where ${tex`a(t)>0`} which is ${tex.block`6t-24>0`} which occurs when ${tex`t>4`}. Therefore we need the times that are in BOTH ${tex`[0,2)\cup (6,\infty)`} AND ${tex`(4,\infty)`}. The times that are in both is the set ${tex`(6,\infty)`}. But that is only the part when the velocity is positive and increasing. We also need the times when velocity is negative and decreasing. Of course velocity is negative when time is in ${tex`(2,6)`}. Velocity is decreasing when ${tex`t < 4`}. We need the times that are in both of these sets, which is just the times ${tex`(2,4)`}. Finally we can conclude that the particle is speeding up when time is in the set ${tex`(2,4)\cup 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