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?
$s(t)=0$

Solution:

(a)

Step 1:
Let $v(t)$ be the velocity of the object at time $t.$
Then, we have
${\begin{array}{rcl}\displaystyle {v(3)}&=&\displaystyle {\lim _{t\rightarrow 3}{\frac {s(t)-s(3)}{t-3}}}\\&&\\&=&\displaystyle {\lim _{t\rightarrow 3}{\frac {-4.9t^{2}+200-(-4.9(9)+200)}{t-3}}}\\&&\\&=&\displaystyle {\lim _{t\rightarrow 3}{\frac {-4.9t^{2}+44.1}{t-3}}.}\end{array}}$

Step 2:
Now, we factor the numerator to get
${\begin{array}{rcl}\displaystyle {v(3)}&=&\displaystyle {\lim _{t\rightarrow 3}{\frac {-4.9t^{2}+44.1}{t-3}}}\\&&\\&=&\displaystyle {\lim _{t\rightarrow 3}{\frac {-4.9(t^{2}-9)}{t-3}}}\\&&\\&=&\displaystyle {\lim _{t\rightarrow 3}{\frac {-4.9(t-3)(t+3)}{(t-3)}}}\\&&\\&=&\displaystyle {\lim _{t\rightarrow 3}-4.9(t+3)}\\&&\\&=&\displaystyle {6(-4.9){\text{ meters/second}}.}\end{array}}$

(b)

Step 1:
First, we need to find the time when the object hits the ground.
This corresponds to $s(t)=0.$
This give us the equation
$-4.9t^{2}+200=0.$
When we solve for $t,$ we get
$t^{2}={\frac {200}{4.9}}.$
Hence, $t=\pm {\sqrt {\frac {200}{4.9}}}.$
Since $t$ represents time, it does not make sense for $t$ 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)     $-9.8{\sqrt {\frac {200}{4.9}}}{\text{ meters/second}}$

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