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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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
${\begin{array}{rcl}\displaystyle {\int _{0}^{10}|-32t+200|~dt}&=&\displaystyle {\int _{0}^{6.25}-32t+200~dt+\int _{6.25}^{10}-(-32t+200)~dt}\\&&\\&=&\displaystyle {\left.(-16t^{2}+200t)\right|_{0}^{6.25}+\left.(16t^{2}-200t)\right|_{6.25}^{10}}\\&&\\&=&\displaystyle {-16(6.25)^{2}+200(6.25)+(16(10)^{2}-200(10))-(16(6.25)^{2}-200(6.25))}\\&&\\&=&\displaystyle {850}.\\\end{array}}$

Final Answer:
The particle travels  $850$   feet.

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