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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle s(t)?}
$v(t)=s'(t)$
2. What is the position of the object when it hits the ground?
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle s(t)=0.}
This give us the equation
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -4.9t^{2}+200=0.}
When we solve for  $t,$  we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle t^{2}={\frac {200}{4.9}}.}
Hence,   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -9.8{\sqrt {\frac {200}{4.9}}}{\text{ meters/second}}.}

Final Answer:
(a)     $6(-4.9){\text{ meters/second}}$
(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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