Difference between revisions of "009A Sample Midterm 3, Problem 2"
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| − | <span class="exam">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. | + | <span class="exam">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. |
| − | <span class="exam">The velocity at time <math style="vertical-align: -1px">t=a</math> seconds is given by: | + | <span class="exam">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> | ::<math>\lim_{t\rightarrow a} \frac{s(t)-s(a)}{t-a}</math> | ||
| − | <span class="exam">(a) Find the velocity of the object when <math style="vertical-align: -1px">t=3</math> | + | <span class="exam">(a) Find the velocity of the object when <math style="vertical-align: -1px">t=3.</math> |
<span class="exam">(b) At what velocity will the object impact the ground? | <span class="exam">(b) At what velocity will the object impact the ground? | ||
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!Foundations: | !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> | + | |'''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> | | <math>v(t)=s'(t)</math> | ||
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!Step 1: | !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> | + | |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 | |Then, we have | ||
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|First, we need to find the time when the object hits the ground. | |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 corresponds to <math style="vertical-align: -5px">s(t)=0.</math> |
|- | |- | ||
|This give us the equation | |This give us the equation | ||
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| <math>-4.9t^2+200=0.</math> | | <math>-4.9t^2+200=0.</math> | ||
|- | |- | ||
| − | |When we solve for <math style="vertical-align: -5px">t,</math> we get | + | |When we solve for <math style="vertical-align: -5px">t,</math> we get |
|- | |- | ||
| <math>t^2=\frac{200}{4.9}.</math> | | <math>t^2=\frac{200}{4.9}.</math> | ||
|- | |- | ||
| − | |Hence, <math style="vertical-align: -18px">t=\pm \sqrt{\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. | + | |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> | + | |Therefore, the object hits the ground at <math style="vertical-align: -18px">t=\sqrt{\frac{200}{4.9}}.</math> |
|} | |} | ||
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|Now, we need the equation for the velocity of the object. | |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. | + | |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, | |Hence, | ||
Revision as of 17:49, 26 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: |
|---|
| 1. What is the relationship between velocity 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)} and position 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)?} |
| 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)} |
| 2. What is the position of the object when it hits the ground? |
| 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} |
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 |
|
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{-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}} |
(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 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}} |