MathAdmin at 00:05, 12 November 2017

2017-11-12T00:05:39Z

MathAdmin: Created page with "The position function $s(t)=-4.9t^2+200$ gives the height (in meters) of an object that has fallen from..."

2017-04-09T18:55:26Z

Created page with "The position function <math style="vertical-align: -5px">s(t)=-4.9t^2+200</math> gives the height (in meters) of an object that has fallen from..."

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The position function  <math style="vertical-align: -5px">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 style="vertical-align: -1px">t=a</math>  seconds is given by:
::<math>\lim_{t\rightarrow a} \frac{s(t)-s(a)}{t-a}</math>

(a) Find the velocity of the object when  <math style="vertical-align: -1px">t=3.</math>

(b) At what velocity will the object impact the ground?
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What is the relationship between velocity  <math style="vertical-align: -5px">v(t)</math>  and position  <math style="vertical-align: -5px">s(t)?</math>
|-
|        <math>v(t)=s'(t)</math>
|-
|'''2.''' What is the position of the object when it hits the ground?
|-
|        <math>s(t)=0</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -5px">v(t)</math>  be the velocity of the object at time  <math style="vertical-align: -1px">t.</math>
|-
|Then, we have
|-
|        <math>\begin{array}{rcl}
\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}}\\
&&\\
& = & \displaystyle{\lim_{t\rightarrow 3} \frac{-4.9t^2+44.1}{t-3}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we factor the numerator to get
|-
|
        <math>\begin{array}{rcl}
\displaystyle{v(3)} & = & \displaystyle{\lim_{t\rightarrow 3} \frac{-4.9t^2+44.1}{t-3}}\\
&&\\
& = & \displaystyle{\lim_{t\rightarrow 3} \frac{-4.9(t^2-9)}{t-3}}\\
&&\\
& = & \displaystyle{\lim_{t\rightarrow 3} \frac{-4.9(t-3)(t+3)}{(t-3)}}\\
&&\\
& = & \displaystyle{\lim_{t\rightarrow 3} -4.9(t+3)}\\
&&\\
& = & \displaystyle{6(-4.9) \text{ meters/second}.}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to find the time when the object hits the ground.
|-
|This corresponds to  <math style="vertical-align: -5px">s(t)=0.</math>
|-
|This give us the equation
|-
|        <math>-4.9t^2+200=0.</math>
|-
|When we solve for  <math style="vertical-align: -5px">t,</math>  we get
|-
|        <math>t^2=\frac{200}{4.9}.</math>
|-
|Hence,   <math style="vertical-align: -18px">t=\pm \sqrt{\frac{200}{4.9}}.</math>
|-
|Since  <math style="vertical-align: 0px">t</math>  represents time, it does not make sense for  <math style="vertical-align: 0px">t</math>  to be negative.
|-
|Therefore, the object hits the ground at  <math style="vertical-align: -18px">t=\sqrt{\frac{200}{4.9}}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we need the equation for the velocity of the object.
|-
|We have  <math style="vertical-align: -5px">v(t)=s'(t)</math>  where  <math style="vertical-align: -5px">v(t)</math>  is the velocity function of the object.
|-
|Hence,
|-
|
        <math>\begin{array}{rcl}
\displaystyle{v(t)} & = & \displaystyle{s'(t)}\\
&&\\
& = & \displaystyle{-9.8t.}
\end{array}</math>
|-
|Therefore, the velocity of the object when it hits the ground is
|-
|        <math>-9.8\sqrt{\frac{200}{4.9}}\text{ meters/second}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>6(-4.9) \text{ meters/second}</math>
|-
|    '''(b)'''     <math>-9.8\sqrt{\frac{200}{4.9}}\text{ meters/second}</math>
|}
[[009A_Sample_Midterm_3|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
-<span class="exam">The position function &nbsp;<math style="vertical-align: -5px">s(t)=-4.9t^2+200</math>&nbsp; gives the height (in meters) of an object that has fallen from a height of 200 meters.
+<span class="exam">Sketch the graph of &nbsp;<math style="vertical-align: -4px">f.</math>&nbsp; At each point of discontinuity, state whether &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is left or right continuous.
-<span class="exam">The velocity at time &nbsp;<math style="vertical-align: -1px">t=a</math>&nbsp; seconds is given by:
+::<math>f(x)=\begin{array}{cc}
-::<math>\lim_{t\rightarrow a} \frac{s(t)-s(a)}{t-a}</math>
+  \Bigg\{ &
+    \begin{array}{cc}
+      x^3+1 & x\leq 0 \\
-<span class="exam">(a) Find the velocity of the object when &nbsp;<math style="vertical-align: -1px">t=3.</math>
+      -x+1 & 0< x< 2 \\
+      -x^2+10x-15 & x\ge 2
-<span class="exam">(b) At what velocity will the object impact the ground?
+    \end{array}
 \end{array}</math>
-|}
+<hr>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A Sample Midterm 3, Problem 2 Solution|'''<u>Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A Sample Midterm 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

009A Sample Midterm 3, Problem 2 - Revision history

MathAdmin at 00:05, 12 November 2017

MathAdmin: Created page with "The position function s(t)=-4.9t^2+200 gives the height (in meters) of an object that has fallen from..."

MathAdmin: Created page with "The position function $s(t)=-4.9t^2+200$ gives the height (in meters) of an object that has fallen from..."