# 009B Sample Midterm 2, Problem 3

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

${\displaystyle v(t)=-32t+200}$

feet per second. Determine the total distance traveled by the particle

from time  ${\displaystyle t=0}$  to time  ${\displaystyle t=10.}$

Foundations:
1. How are the velocity function  ${\displaystyle v(t)}$  and the position function  ${\displaystyle s(t)}$  related?

They are related by the equation  ${\displaystyle v(t)=s'(t).}$

2. If we calculate  ${\displaystyle \int _{a}^{b}v(t)~dt,}$  what are we calculating?

We are calculating  ${\displaystyle s(b)-s(a).}$

This is the displacement of the particle from  ${\displaystyle t=a}$  to  ${\displaystyle t=b.}$

3. If we calculate  ${\displaystyle \int _{a}^{b}|v(t)|~dt,}$  what are we calculating?

We are calculating the total distance traveled by the particle from  ${\displaystyle t=a}$  to  ${\displaystyle t=b.}$

Solution:

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

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

The particle travels  ${\displaystyle 850}$   feet.