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 $t=10.$

ExpandFoundations:
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:

ExpandStep 1:
To calculate the total distance the particle traveled from $t=0$ to $t=10,$
we need to calculate
$\int _{0}^{10}|v(t)|~dt=\int _{0}^{10}|-32t+200|~dt.$

ExpandStep 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 $[6.25,10].$

ExpandStep 3:
Therefore, we get
${\begin{array}{rcl}\displaystyle {\int _{0}^{10}|-32t+200|~dt}&=&\displaystyle {\int _{0}^{6.25}-32t+200~dt+\int _{6.25}^{10}-(-32t+200)~dt}\\&&\\&=&\displaystyle {\left.(-16t^{2}+200t)\right|_{0}^{6.25}+\left.(16t^{2}-200t)\right|_{6.25}^{10}}\\&&\\&=&\displaystyle {-16(6.25)^{2}+200(6.25)+(16(10)^{2}-200(10))-(16(6.25)^{2}-200(6.25))}\\&&\\&=&\displaystyle {850}.\\\end{array}}$

ExpandFinal Answer:
The particle travels $850$ feet.

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