Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 4

Suppose that a particle moves according to a Law of Motion

$\displaystyle f(t) = \frac{t}{1 + t^2}$, where $t$ is measured in
seconds and $s$ in feet.

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




$
\begin{equation}
\begin{aligned}

\text{velocity } =& s'(t) = \frac{ds}{dt}
\\
\\
& \text{Using Quotient Rule}
\\
\\
v(t) =& \frac{\displaystyle (1 + t^2) \frac{d}{dt} (t) - t \frac{d}{dt} (1 + t^2)}{(1 + t^2)^2}
\\
\\
v(t) =& \frac{(1 + t^2)(1) - t (2t) }{(1 + t^2)^2}
\\
\\
v(t) =& \frac{1 + t^2 - 2t^2}{(1 + t^2)^2}
\\
\\
v(t) =& \frac{-t^2 + 1}{(1 + t^2)^2}

\end{aligned}
\end{equation}
$


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


$
\begin{equation}
\begin{aligned}

\text{The velocity after $3 s$ is } v(3) =& \frac{-(3)^2 + 1}{(1 + (3)^2)^2}
\\
\\
v(3) =& -0.08 ft/s

\end{aligned}
\end{equation}
$


c.) When is the velocity at rest?

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


$
\begin{equation}
\begin{aligned}

0 =& \frac{-t^2 + 1}{(1 + t^2)^2}
\\
\\
t^2 =& 1
\\
\\
t =& \pm \sqrt{1}
\\
\\
t =& 1 \text{ and } t = -1
\end{aligned}
\end{equation}
$


The required time is $t = 1$ since the position function is defined only for positive values of $t$.

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

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


$
\begin{equation}
\begin{aligned}

& \frac{-t^2 + 1}{(1 + t^2)^2} > 0

\end{aligned}
\end{equation}
$


Assume $\displaystyle \frac{-t^2 + 1}{(1 + t^2)^2}$ we have $t = 1$


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

(i) $0 \leq t \leq 1$

Let's assume $t = 0.5: \displaystyle \frac{-(0.5)^2 + 1}{(1 + (0.5)^2)^2}= 0.48 > 0$

(ii) $t > 1$

Let's assume $\displaystyle t = 3: \frac{-(3)^2 + 1}{(1 + (3)^2)^2} = -0.08 < 0$

Therefore, we can conclude that the particle is speeding up at the
interval $0 \leq t \leq 1$. However, the particle 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
$t = 1$, we take the distance that the particle traveled on the intervals $(0,1)$ and $(1,8)$


$
\begin{equation}
\begin{aligned}

\text{Total distance } =& |f(1) - f(0)| + f(8) - f(1) | ; f(t) = \frac{t}{1 + t^2}
\\
\\
=& |0.5 - 0 | + \left|\frac{8}{65} - \frac{1}{2} \right|
\\
\\
=& 0.5 + \left| - \frac{49}{130} \right|
\\
\\
=& 0.5 + \frac{49}{130}
\\
\\
=& 0.8769 ft

\end{aligned}
\end{equation}
$



f.) Illustrate the motion of the particle.







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

Using Quotient Rule and Chain Rule,


$
\begin{equation}
\begin{aligned}

\text{acceleration } =& v'(t) = \frac{dv}{dt}
\\
\\
a(t) =& \frac{\displaystyle (1 + t^2)^2 \cdot \frac{d}{dt} (-t^2 + 1) - (-t^2 + 1) \cdot \frac{d}{dt} (1 + t^2)^2 }{[(1 + t^2)^2]^2 }
\\
\\
a(t) =& \frac{(1 + t^2) \cdot (-2t) - (-t^2 + 1) [2 (1 + t^2) \cdot (2t)] }{[(1 + t^2)^2]^2}
\\
\\
a(t) =& \frac{\cancel{(1 + t^2)} [-2t(1 + t^2) - 4t (-t^2 + 1)] }{(1 + t^2)^{\cancel{4}}}
\\
\\
a(t) =& \frac{-2t - 2t^3 + 4t^3 - 4t}{(1 + t^2)^3}
\\
\\
a(t) =& \frac{2t^3 - 6t}{(1 + t^2)^3}
\\
\\
\text{Acceleration at } t = 3,
\\
\\
a(3) =& \frac{2(3)^3 - 6 (3)}{(1 + (3)^2)^3}
\\
\\
a(3) =& 0.036 ft/s^2
\end{aligned}
\end{equation}
$


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







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

Based from the graph, the particle is speeding up when the velocity
and acceleration have the same sign (either positive or negative). Notice in the curve that the acceleration changes direction in between $1 < t < 2$ where $a(t) = 0$ so,


$
\begin{equation}
\begin{aligned}

0 =& \frac{2t^3 - 6t}{(1 + t^2)^3}
\\
\\
2t^3 =& 6t
\\
\\
t^2 =& 3
\\
\\
t =& \pm \sqrt{3}
\end{aligned}
\end{equation}
$


Hence, we can conclude that the particle is speeding up at interval

$1 < t < \sqrt{3}$

On the other hand, the particle is slowing down when the velocity and acceleration have opposite sign that is $0 \leq t \leq 1$ and $\sqrt{3} \leq t \leq 8$.