009A Sample Midterm 3, Problem 2

The position function $s(t)=-4.9t^{2}+200$ gives the height (in meters) of an object that has fallen from a height of 200 meters.

The velocity at time $t=a$ seconds is given by:

$\lim _{t\rightarrow a}{\frac {s(t)-s(a)}{t-a}}$

(a) Find the velocity of the object when $t=3$

(b) At what velocity will the object impact the ground?

Foundations:
1. What is the relationship between velocity $v(t)$ and position $s(t)?$
$v(t)=s'(t)$
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 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 $t={\sqrt {\frac {200}{4.9}}}.$

Step 2:
Now, we need the equation for the velocity of the object.
We have $v(t)=s'(t)$ where $v(t)$ is the velocity function of the object.
Hence,
${\begin{array}{rcl}\displaystyle {v(t)}&=&\displaystyle {s'(t)}\\&&\\&=&\displaystyle {-9.8t.}\end{array}}$
Therefore, the velocity of the object when it hits the ground is
$-9.8{\sqrt {\frac {200}{4.9}}}{\text{ meters/second}}.$

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

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