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
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 \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 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 850} feet.

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