007A Sample Midterm 3, Problem 5 Detailed Solution

At time  $t,$  the position of a body moving along the  $s-$axis is given by  $s=t^{3}-6t^{2}+9t$ (in meters and seconds).

(a)  Find the times when the velocity of the body is equal to  $0.$

(b)  Find the body's acceleration each time the velocity is  $0.$

(c)  Find the total distance traveled by the body from time  $t=0$  second to  $t=2$  seconds.

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Background Information:
1. If  $s$  is the position function of an object and
$v$  is the velocity function of that same object,
then   $v=s'.$
2. If  $v$  is the velocity function of an object and
$a$  is the acceleration function of that same object,
then  $a=v'.$

Solution:

(a)

Step 1:
We begin by noticing that we plug in  $x=0$  into
${\frac {x}{3-{\sqrt {9-x}}}},$
we get   ${\frac {0}{0}}.$

Step 2:
Now, we multiply the numerator and denominator by the conjugate of the denominator.
Hence, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {x}{3-{\sqrt {9-x}}}}}&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {x}{3-{\sqrt {9-x}}}}{\frac {(3+{\sqrt {9-x}})}{(3+{\sqrt {9-x}})}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {x(3+{\sqrt {9-x}})}{9-(9-x)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {x(3+{\sqrt {9-x}})}{x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {3+{\sqrt {9-x}}}{1}}}\\&&\\&=&\displaystyle {\frac {3+{\sqrt {9}}}{1}}\\&&\\&=&\displaystyle {\frac {6}{1}}\\&&\\&=&\displaystyle {6.}\end{array}}$

(b)

Step 1:
We proceed using L'Hôpital's Rule. So, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \pi }{\frac {\sin(x)}{\pi -x}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \pi }{\frac {\cos(x)}{-1}}.}\end{array}}$

Step 2:
Now, we plug in  $x=\pi$  to get
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \pi }{\frac {\sin(x)}{\pi -x}}}&=&\displaystyle {\frac {\cos(\pi )}{-1}}\\&&\\&=&\displaystyle {\frac {-1}{-1}}\\&&\\&=&\displaystyle {1.}\end{array}}$

(c)

Step 1:
We begin by factoring the numerator and denominator. We have
$\lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}\,=\,\lim _{x\rightarrow -2}{\frac {(x+2)(x-3)}{(x+2)(x^{2}-2x+4)}}.$
So, we can cancel  $x+2$  in the numerator and denominator. Thus, we have
$\lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}\,=\,\lim _{x\rightarrow -2}{\frac {x-3}{x^{2}-2x+4}}.$

Step 2:
Now, we can just plug in  $x=-2$  to get
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -2}{\frac {x^{2}-x-6}{x^{3}+8}}}&=&\displaystyle {\frac {-2-3}{(-2)^{2}-2(-2)+4}}\\&&\\&=&\displaystyle {-{\frac {5}{12}}.}\end{array}}$

Final Answer:
(a)
(b)
(c)

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