# 009B Sample Midterm 2, Problem 3

A particle moves along a straight line with velocity given by:

$v(t)=-32t+200$ feet per second. Determine the total distance traveled by the particle

from time  $t=0$ to time  $t=10.$ Foundations:
1. How are the velocity function  $v(t)$ and the position function  $s(t)$ related?

They are related by the equation  $v(t)=s'(t).$ 2. If we calculate  $\int _{a}^{b}v(t)~dt,$ what are we calculating?

We are calculating  $s(b)-s(a).$ This is the displacement of the particle from  $t=a$ to  $t=b.$ 3. If we calculate  $\int _{a}^{b}|v(t)|~dt,$ what are we calculating?

We are calculating the total distance traveled by the particle from  $t=a$ to  $t=b.$ Solution:

Step 1:
To calculate the total distance the particle traveled from  $t=0$ to  $t=10,$ we need to calculate
$\int _{0}^{10}|v(t)|~dt=\int _{0}^{10}|-32t+200|~dt.$ Step 2:
We need to figure out when  $-32t+200$ is positive and negative in the interval  $[0,10].$ We set
$-32t+200=0$ and solve for  $t.$ We get
$t=6.25.$ Then, we use test points to see that  $-32t+200$ is positive from  $[0,6.25]$ and negative from  $[6.25,10].$ Step 3:
Therefore, we get

${\begin{array}{rcl}\displaystyle {\int _{0}^{10}|-32t+200|~dt}&=&\displaystyle {\int _{0}^{6.25}-32t+200~dt+\int _{6.25}^{10}-(-32t+200)~dt}\\&&\\&=&\displaystyle {\left.(-16t^{2}+200t)\right|_{0}^{6.25}+\left.(16t^{2}-200t)\right|_{6.25}^{10}}\\&&\\&=&\displaystyle {-16(6.25)^{2}+200(6.25)+(16(10)^{2}-200(10))-(16(6.25)^{2}-200(6.25))}\\&&\\&=&\displaystyle {850}.\\\end{array}}$ The particle travels  $850$ feet.