# 007A Sample Midterm 3, Problem 5 Detailed Solution

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

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

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

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

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

Solution:

(a)

Step 1:
First, we need to find the velocity function of this body.
By the Power Rule, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {v}&=&\displaystyle {s'}\\&&\\&=&\displaystyle {(t^{3}-6t^{2}+9t)'}\\&&\\&=&\displaystyle {3t^{2}-12t+9.}\end{array}}}$
Step 2:
Now, we set the velocity function equal to  ${\displaystyle 0}$  and solve.
Hence, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {3t^{2}-12t+9}\\&&\\&=&\displaystyle {3(t^{2}-4t+3)}\\&&\\&=&\displaystyle {3(t-1)(t-3).}\end{array}}}$
So, the two solutions are  ${\displaystyle t=1}$  and  ${\displaystyle t=3.}$
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

${\displaystyle {\begin{array}{rcl}\displaystyle {a}&=&\displaystyle {v'}\\&&\\&=&\displaystyle {(3t^{2}-12t+9)'}\\&&\\&=&\displaystyle {6t-12.}\end{array}}}$

Step 2:
Now, we plug in  ${\displaystyle t=1}$  and  ${\displaystyle t=3.}$
When  ${\displaystyle t=1,}$  we get

${\displaystyle {\begin{array}{rcl}\displaystyle {a}&=&\displaystyle {6-12}\\&&\\&=&\displaystyle {-6~{\frac {{\text{m}}^{2}}{\text{s}}}.}\end{array}}}$

When  ${\displaystyle t=3,}$  we get

${\displaystyle {\begin{array}{rcl}\displaystyle {a}&=&\displaystyle {6(3)-12}\\&&\\&=&\displaystyle {6~{\frac {{\text{m}}^{2}}{\text{s}}}.}\end{array}}}$

(c)

Step 1:
Since the velocity is  ${\displaystyle 0}$  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
${\displaystyle s(0)=0,~s(1)=4,~s(2)=2.}$
Step 2:
Hence, the total distance the body traveled is

${\displaystyle {\begin{array}{rcl}\displaystyle {[s(1)-s(0)]+[s(1)-s(2)]}&=&\displaystyle {[4-0]+[4-2]}\\&&\\&=&\displaystyle {6{\text{ meters}}.}\end{array}}}$

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