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| − | <span class="exam">The position function <math>s(t)=-4.9t^2+200</math> gives the height (in meters) of an object that has fallen from a height of 200 meters. The velocity at time <math>t=a</math> seconds is given by: | + | <span class="exam">Sketch the graph of <math style="vertical-align: -4px">f.</math> At each point of discontinuity, state whether <math style="vertical-align: -4px">f</math> is left or right continuous. |
| − | ::::::<math>\lim_{t\rightarrow a} \frac{s(t)-s(a)}{t-a}</math>
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| − | | + | ::<math>f(x)=\begin{array}{cc} |
| − | ::<span class="exam">a) Find the velocity of the object when <math>t=3</math> | + | \Bigg\{ & |
| − | ::<span class="exam">b) At what velocity will the object impact the ground?
| + | \begin{array}{cc} |
| − | | + | x^3+1 & x\leq 0 \\ |
| − | | + | -x+1 & 0< x< 2 \\ |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | -x^2+10x-15 & x\ge 2 |
| − | !Foundations:
| + | \end{array} |
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| − | |}
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| − | '''Solution:'''
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |Let <math>v(t)</math> be the velocity of the object at time <math>t.</math>
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| − | |Then, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{v(3)} & = & \displaystyle{\lim_{t\rightarrow 3} \frac{s(t)-s(3)}{t-3}}\\ | |
| − | &&\\ | |
| − | & = & \displaystyle{\lim_{t\rightarrow 3} \frac{-4.9t^2+200-(-4.9(9)+200)}{t-3}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{t\rightarrow 3} \frac{-4.9t^2+44.1}{t-3}.}
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| | \end{array}</math> | | \end{array}</math> |
| − | |}
| + | <hr> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Midterm 3, Problem 2 Solution|'''<u>Solution</u>''']] |
| − | !Step 2:
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| − | |-
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| − | |Now, we factor the numerator to get | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{v(3)} & = & \displaystyle{\lim_{t\rightarrow 3} \frac{-4.9t^2+44.1}{t-3}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{t\rightarrow 3} \frac{-4.9(t^2-9)}{t-3}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{t\rightarrow 3} \frac{-4.9(t-3)(t+3)}{(t-3)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{t\rightarrow 3} -4.9(t+3)}\\
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| − | &&\\
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| − | & = & \displaystyle{6(-4.9) \text{ meters/second}.}
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| − | \end{array}</math>
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| − | |}
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we need to find the time when the object hits the ground.
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| − | |This corresponds to <math>s(t)=0.</math>
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| − | |This give us the equation
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| − | | <math>-4.9t^2+200=0.</math>
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| − | |When we solve for <math>t,</math> we get
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| − | | <math>t^2=\frac{200}{4.9}.</math>
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| − | |-
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| − | |Hence, <math>t=\pm \sqrt{\frac{200}{4.9}}.</math>
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| − | |Since <math>t</math> represents time, it does not make sense for <math>t</math> to be negative.
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| − | |-
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| − | |Therefore, the object hits the ground at <math>t=\sqrt{\frac{200}{4.9}}.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Midterm 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Step 2:
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| − | |- | |
| − | |Now, we need the equation for the velocity of the object.
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| − | |We have <math>v(t)=s'(t)</math> where <math>v(t)</math> is the velocity function of the object.
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| − | |Hence,
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{v(t)} & = & \displaystyle{s'(t)}\\
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| − | &&\\
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| − | & = & \displaystyle{-9.8t.}
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| − | \end{array}</math>
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| − | |-
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| − | |Therefore, the velocity of the object when it hits the ground is
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| − | |-
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| − | | <math>-9.8\sqrt{\frac{200}{4.9}}\text{ meters/second}.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' <math>6(-4.9) \text{ meters/second}</math>
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| − | |-
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| − | | '''(b)''' <math>-9.8\sqrt{\frac{200}{4.9}}\text{ meters/second}</math>
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| − | |}
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| | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
