# 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.

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

${\begin{array}{rcl}\displaystyle {[s(1)-s(0)]+[s(1)-s(2)]}&=&\displaystyle {[4-0]+[4-2]}\\&&\\&=&\displaystyle {6{\text{ meters}}.}\end{array}}$ (a)   $1{\text{ second, }}3{\text{ seconds}}$ (b)   $-6~{\frac {{\text{m}}^{2}}{\text{s}}},~6~{\frac {{\text{m}}^{2}}{\text{s}}}$ (c)   $6{\text{ meters}}$ 