Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 1

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Suppose that a particle moves according to a Law of Motion

$s = f(t) = t^3 - 12t^2 + 36t, t \geq 0$, where $t$ is measured in seconds and $s$ in feet.

a.) Determine the velocity at time $t$.

\begin{aligned} \text{velocity } = s'(t) =& \frac{d}{dt} (t^3) - 12 \frac{d}{dt} (t^2) + 36 \frac{d}{dt} (t) \\ \\ =& 3t^2 - 12 (2t) + 36 (1) \\ \\ =& 3t^2 - 24t + 36 \end{aligned}

b.) What is the velocity after $3 s$?

\begin{aligned} \text{The velocity after 3 s is } v(3) =& 3 (3)^2 - 24(3) + 36 \\ \\ v(3) =& -9 ft/s \end{aligned}

c.) When is the particle at rest?

The particle is at rest when $v(t) = 0$

$0 = 3t^2 - 24t + 36$

$0 = 3(t^2 - 8t + 12)$

$0 = (t - 6)(t - 2)$

Solving for $t$,

$t = 6$ and $t = 2$

The particle is at rest at $t = 6 s$ and $t = 2 s$

d.) When is the particle moving in the positive direction?

The particle is moving in the positive direction when $v(t) > 0$

\begin{aligned} & 3t^2 - 24t + 36 > 0 \\ \\ & 3(t^2 - 8t + 12) > 0 \\ \\ & 3(t - 2) (t - 6) > 0 \end{aligned}

Assume $3(t - 2) (t - 6) = 0$ we have $t = 2$ and $t = 6$.

Dividing the interval $t \geq 0$ into three parts we have,

(i) $t > 6$

Let's assume $t = 7: 3(7)^2 - 24(7) + 36 = 15 > 0$

(ii) $2 < t < 6$

Let's assume $t = 4: 3(4)^2 - 24(4) + 36 = -12 < 0$

(iii) $0 \leq t < 2$

$t = 1: 3(1)^2 - 24(1) + 36 = 15 > 0$

Therefore, we can conclude that the particle is speeding up at the interval $0 \leq t < 2$ and $t > 6$. However, it is moving in the positive direction.

e.) Find the total distance traveled during the first $8 s$.

Since we know that the particle starts at and changes direction at $2$ and $6$, we take the distance it traveled on the intervals $(0,2), (2,6)$ and $(6,8)$

\begin{aligned} \text{Total distance that the particle covers for 8s } =& |f(2) - f(0)| + |f(6) - f(2)| + |f(8) - f(6)| ; f(t) = t^3 - 12t^2 + 36t \\ \\ =& |32 - 0| + |0 - 32| + |32 - 0| \\ \\ =& 32 + 32 + 32 \\ \\ =& 96 ft \end{aligned}

f.) Illustrate the motion of the particle.

g.) Find the acceleration at time $t$ and after $3 s$.

\begin{aligned} \text{acceleration } =& v'(t) = \frac{dv}{dt} \\ \\ =& 3 \frac{d}{dt} (t^2) - 24 \frac{d}{dt} (t) + \frac{d}{dt} (36) \\ \\ =& 3 (2t) - 24(1) + 0 \\ \\ =& 6t - 24 \\ \\ \text{acceleration at } t =& 3, \\ \\ a(3) =& 6 (3) - 24 \\ \\ a(3) =& -6 ft/s^2 \end{aligned}

h.) Graph the position, velocity and acceleration functions $0 \leq t \leq 8$.

i.) When is the particle speeding up? When is it showing down?

Based from the graph, the particle is speeding up when the velocity and acceleration have the same sign (either positive or negative) that is at intervals $2 < t < 4$ and $t > 6$. On the other hand, the particle is slowing down when the velocity and acceleration have opposite sign, that is at intervals $0 < t < 2$ and $4 < t < 6$.