205.o6.03 Find the body's displacement and average velocity

[karush]

$\tiny{205.o6.03}$
$\textsf{A body on a coordinate line such that it has a position}\\$
$\textsf{$\displaystyle s=f(t)=\frac{16}{t^2}-\frac{4}{t}$
on the interval $\displaystyle 1\le t \le 4$
with $s$ in meters and $t$ in seconds }$
$\textsf{a. Find the body's displacement and average vellocity
for the given time interval.}\\$
\begin{align}
\displaystyle
s(4)-s(1)&=\int_{1}^{4}
\left(\frac{16}{t^2}-\frac{4}{t}\right) \,dt \\
&=\left[-4\ln\left({\left| t \right|}\right)-\frac{16}{t}\right]_1^4 \\
&\approx-9.545-(-16)\\
&\approx6.4548 m \textit{ to the right}
\end{align}
$\textsf{b. Find the body's speed and acceration at the endpoints of the interval.}\\$

$\textsf{c. When if ever, during the interval does the body change direction? }$

 

[MarkFL]

Displacement is simply the change in position:

\(\displaystyle \Delta s=s(4)-s(1)=\left[\frac{4}{4}\left(\frac{4}{4}-1\right)\right]-\left[\frac{4}{1}\left(\frac{4}{1}-1\right)\right]=-12\text{ m}\)

Hence, the average velocity is the displacement divided by the change in time:

\(\displaystyle \overline{v}=\frac{\Delta s}{\Delta t}=\frac{-12\text{ m}}{3\text{ s}}=-4\,\frac{\text{m}}{\text{s}}\)

Now, for part b), you will need to apply the calculus...

\(\displaystyle v(t)=s'(t)\) and \(\displaystyle a(t)=v'(t)\)

For part c), which function do we analyze for change in direction? Displacement, velocity, or acceleration?

 

[karush]

$\textsf{(a) Find the body's displacement and average vellocity
for the given time interval.}\\$
\begin{align}
\displaystyle
\Delta s=s(4)-s(1)&=\left[\frac{4}{4}\left(\frac{4}{4}-1\right)\right]-\left[\frac{4}{1}\left(\frac{4}{1}-1\right)\right]=-12\text{ m}\\
\\
v(t)=s'(t)&=\dfrac{4}{t^2}-\dfrac{32}{t^3}\\
\\
\Delta s &=\int_{1}^{4} v(t) \,dt = -12 \ m\\
\end{align}
$\textit{so the integral of $s'(v)$ or $v(t)$ will also give us displacement?}$

 

[MarkFL]

To find the velocity you will need to differentiate the displacement. Then, if you desire to find the overall change in displacement from the velocity, you would need to integrate...however, this would bring you back to where you were had you simply used the given displacement function in the first place.

If you had only been given the velocity function, then it would make sense to integrate in order to find the displacement. But, you were given the displacement function, so you only need to use that given function to compute a change in displacement.

 

[karush]

$\textit{so s(t) is the displacement equation?}$

$\textsf{(b) Find the body's speed and acceration at the endpoints of the interval.}\\$
\begin{align}
\displaystyle
v(t)=s'(t)&=\dfrac{4}{t^2}-\dfrac{32}{t^3}\\
v(1)&=\dfrac{4}{(1)^2}-\dfrac{32}{(1)^3}=4-32=-28 \\
v(4)&=\dfrac{4}{(4)^2}-\dfrac{32}{(4)^3}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{2} \\
\end{align}
$\textit{if $v(t)$ is speed the why are we getting negative values?}$

 

[MarkFL]

$\textit{so s(t) is the displacement equation?}$

$\textsf{(b) Find the body's speed and acceration at the endpoints of the interval.}\\$
\begin{align}
\displaystyle
v(t)=s'(t)&=\dfrac{4}{t^2}-\dfrac{32}{t^3}\\
v(1)&=\dfrac{4}{(1)^2}-\dfrac{32}{(1)^3}=4-32=-28 \\
v(4)&=\dfrac{4}{(4)^2}-\dfrac{32}{(4)^3}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{2} \\
\end{align}
$\textit{if $v(t)$ is speed the why are we getting negative values?}$

Check your arithmetic for $v(4)$. :D

Velocity is a vector, with a speed and direction. In one-dimensional motion, the sign of the function determines the direction...negative is to the left while positive is to the right. Speed is simply the magnitude of the velocity (the absolute value), and has no direction, it is a scalar.

For example, if I tell you I am going 90 mph on the highway, I have told you my speed. But if I tell you I am traveling south at 90 mph on the highway, then I have told you my velocity.

 

[karush]

\begin{align}
\displaystyle
v(t)=s'(t)&=\dfrac{4}{t^2}-\dfrac{32}{t^3}\\
v(1)&=\dfrac{4}{(1)^2}-\dfrac{32}{(1)^3}=4-32= \left| -28 \right| =28 \ m/s \\
v(4)&=\dfrac{4}{(4)^2}-\dfrac{32}{(4)^3}=\frac{1}{4}-\frac{1}{2}= \left| -\frac{1}{4} \right| =0.25 \ m/s \\
\end{align}

better!

for (c) I got $t=12$
but that's not in the interval...

 

[MarkFL]

It would be better to write the speeds this way:

\(\displaystyle \left|v(4)\right|=\cdots=\frac{1}{4}\,\frac{\text{m}}{\text{s}}\)

For c) you want to write:

\(\displaystyle v(t)=0\)

\(\displaystyle \frac{4}{t^2}-\frac{32}{t^3}=0\)

Multiply through by \(\displaystyle \frac{t^3}{4}\ne0\)

\(\displaystyle t-8=0\)

\(\displaystyle t=8\)

Thus, we know the direction does not change on the given interval.

 

[karush]

ok, I did $a(t)=0$

I do some more of these ... too rusty..