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  $t={\sqrt {\frac {200}{4.9}}}.$

Step 2:
Now, we calculate the velocity.
We have
${\begin{array}{rcl}\displaystyle {v{\bigg (}{\sqrt {\frac {200}{4.9}}}{\bigg )}}&=&\displaystyle {\lim _{t\rightarrow {\sqrt {\frac {200}{4.9}}}}{\frac {s(t)-s{\bigg (}{\sqrt {\frac {200}{4.9}}}{\bigg )}}{t-{\sqrt {\frac {200}{4.9}}}}}}\\&&\\&=&\displaystyle {\lim _{t\rightarrow {\sqrt {\frac {200}{4.9}}}}{\frac {-4.9t^{2}+200-(-4.9{\big (}{\frac {200}{4.9}}{\big )}+200)}{t-{\sqrt {\frac {200}{4.9}}}}}}\\&&\\&=&\displaystyle {\lim _{t\rightarrow {\sqrt {\frac {200}{4.9}}}}{\frac {-4.9t^{2}+200}{t-{\sqrt {\frac {200}{4.9}}}}}}\\&&\\&=&\displaystyle {\lim _{t\rightarrow {\sqrt {\frac {200}{4.9}}}}{\frac {-4.9{\big (}t-{\sqrt {\frac {200}{4.9}}}{\big )}{\big (}t+{\sqrt {\frac {200}{4.9}}}{\big )}}{t-{\sqrt {\frac {200}{4.9}}}}}}\\&&\\&=&\displaystyle {\lim _{t\rightarrow {\sqrt {\frac {200}{4.9}}}}-4.9{\bigg (}t+{\sqrt {\frac {200}{4.9}}}{\bigg )}}\\&&\\&=&\displaystyle {-4.9{\bigg (}{\sqrt {\frac {200}{4.9}}}+{\sqrt {\frac {200}{4.9}}}{\bigg )}}\\&&\\&=&\displaystyle {-9.8{\sqrt {\frac {200}{4.9}}}{\text{ meters/second}}.}\end{array}}$

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

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