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:

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:

Step 2:

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

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