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:
First, we need to find the velocity function of this body.
By the Power Rule, we have
${\begin{array}{rcl}\displaystyle {v}&=&\displaystyle {s'}\\&&\\&=&\displaystyle {(t^{3}-6t^{2}+9t)'}\\&&\\&=&\displaystyle {3t^{2}-12t+9.}\end{array}}$

Step 2:
Now, we set the velocity function equal to  $0$  and solve.
Hence, we have
${\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {3t^{2}-12t+9}\\&&\\&=&\displaystyle {3(t^{2}-4t+3)}\\&&\\&=&\displaystyle {3(t-1)(t-3).}\end{array}}$
So, the two solutions are  $t=1$  and  $t=3.$
Therefore, the velocity is zero at 1 second and 3 seconds.

(b)

Step 1:
First, we need to find the acceleration function of this body.
Using the Power Rule again, we have
${\begin{array}{rcl}\displaystyle {a}&=&\displaystyle {v'}\\&&\\&=&\displaystyle {(3t^{2}-12t+9)'}\\&&\\&=&\displaystyle {6t-12.}\end{array}}$

Step 2:
Now, we plug in  $t=1$  and  $t=3.$
When  $t=1,$  we get
${\begin{array}{rcl}\displaystyle {a}&=&\displaystyle {6-12}\\&&\\&=&\displaystyle {-6~{\frac {{\text{m}}^{2}}{\text{s}}}.}\end{array}}$
When  $t=3,$  we get
${\begin{array}{rcl}\displaystyle {a}&=&\displaystyle {6(3)-12}\\&&\\&=&\displaystyle {6~{\frac {{\text{m}}^{2}}{\text{s}}}.}\end{array}}$

(c)

Step 1:
Since the velocity is  $0$  at 1 second,
we need to consider the position of this body at 0, 1, and 2 seconds.
Plugging these values into the position function, we get
$s(0)=0,~s(1)=4,~s(2)=2.$

Step 2:
Hence, the total distance the body traveled is
${\begin{array}{rcl}\displaystyle {[s(1)-s(0)]+[s(1)-s(2)]}&=&\displaystyle {[4-0]+[4-2]}\\&&\\&=&\displaystyle {6{\text{ meters}}.}\end{array}}$

Final Answer:
(a)   $1{\text{ second, }}3{\text{ seconds}}$
(b)   $-6~{\frac {{\text{m}}^{2}}{\text{s}}},~6~{\frac {{\text{m}}^{2}}{\text{s}}}$
(c)   $6{\text{ meters}}$

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