Difference between revisions of "009A Sample Midterm 3, Problem 2"

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
|
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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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|&nbsp; &nbsp; &nbsp; &nbsp; <math>-4.9t^2+200=0.</math>
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|-
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|When we solve for <math>t,</math> we get
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>t^2=\frac{200}{4.9}.</math>
 
|-
 
|-
|
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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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|Therefore, the object hits the ground at <math>t=\sqrt{\frac{200}{4.9}}.</math>
 
|}
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|  
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|Now, we need the equation for the velocity of the object. 
 +
|-
 +
|We have <math>v(t)=s'(t)</math> where <math>v(t)</math> is the velocity function of the object.
 +
|-
 +
|Hence,
 
|-
 
|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp; <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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|Therefore, the velocity of the object when it hits the ground is
 
|-
 
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>-9.8\sqrt{\frac{200}{4.9}}\text{ meters/second}.</math>
 
|}
 
|}
  
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>6(-4.9) \text{ meters/second}</math>  
 
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>6(-4.9) \text{ meters/second}</math>  
 
|-
 
|-
|'''(b)'''  
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>-9.8\sqrt{\frac{200}{4.9}}\text{ meters/second}</math>
 
|}
 
|}
 
[[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 14:19, 17 February 2017

The position function 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 s(t)=-4.9t^2+200} gives the height (in meters) of an object that has fallen from a height of 200 meters. The velocity at 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=a} seconds is given by:

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 \lim_{t\rightarrow a} \frac{s(t)-s(a)}{t-a}}


a) Find the velocity of the object when 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=3}
b) At what velocity will the object impact the ground?


Foundations:  


Solution:

(a)

Step 1:  
Let 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 v(t)} be the velocity of the object at 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.}
Then, we have
        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{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}}
Step 2:  
Now, we factor the numerator to get

       

(b)

Step 1:  
First, we need to find the time when the object hits the ground.
This corresponds to 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 s(t)=0.}
This give us the equation
        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 -4.9t^2+200=0.}
When we solve for 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,} 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 t^2=\frac{200}{4.9}.}
Hence, 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=\pm \sqrt{\frac{200}{4.9}}.}
Since 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} represents time, it does not make sense for 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} to be negative.
Therefore, the object hits the ground at 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=\sqrt{\frac{200}{4.9}}.}
Step 2:  
Now, we need the equation for the velocity of the object.
We have 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 v(t)=s'(t)} where 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 v(t)} is the velocity function of the object.
Hence,

        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{v(t)} & = & \displaystyle{s'(t)}\\ &&\\ & = & \displaystyle{-9.8t.} \end{array}}

Therefore, the velocity of the object when it hits the ground is
        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 -9.8\sqrt{\frac{200}{4.9}}\text{ meters/second}.}


Final Answer:  
    (a)     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 6(-4.9) \text{ meters/second}}
    (b)     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 -9.8\sqrt{\frac{200}{4.9}}\text{ meters/second}}

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