Certainly! Let's carefully address each part of this problem step-by-step:
Problem Statement:
An object is thrown upward from a height of 6 feet at a velocity of 63 feet per second. Given [tex]\( t_1 = 0 \)[/tex] and [tex]\( t_2 = 3 \)[/tex]:
(a) Use the position equation [tex]\( s = -16 t^2 + v_0 t + s_0 \)[/tex] to write a function that represents the situation. Use a graphing utility to graph the function.
[tex]\[
s(t) =
\][/tex]
Let's define the variables:
- Initial velocity ([tex]\(v_0\)[/tex]) = 63 feet per second
- Initial position ([tex]\(s_0\)[/tex]) = 6 feet
Using the equation [tex]\( s(t) = -16 t^2 + v_0 t + s_0 \)[/tex]:
[tex]\[
s(t) = -16 t^2 + 63 t + 6
\][/tex]
This is the position function [tex]\( s(t) \)[/tex].
(b) Find the average rate of change (in [tex]\( ft/sec \)[/tex] ) of the function from [tex]\( t_1 \)[/tex] to [tex]\( t_2 \)[/tex].
Average rate of change is calculated as:
[tex]\[
\frac{s(t_2) - s(t_1)}{t_2 - t_1}
\][/tex]
We need the position at [tex]\( t_1 = 0 \)[/tex] and [tex]\( t_2 = 3 \)[/tex]:
[tex]\[
s(t_1) = s(0) = -16(0)^2 + 63(0) + 6 = 6
\][/tex]
[tex]\[
s(t_2) = s(3) = -16(3)^2 + 63(3) + 6 = -16(9) + 189 + 6 = -144 + 189 + 6 = 51
\][/tex]
Thus, the average rate of change is:
[tex]\[
\frac{s(3) - s(0)}{3 - 0} = \frac{51 - 6}{3} = \frac{45}{3} = 15 \, ft/sec
\][/tex]
Average rate of change = [tex]\( 15 \, ft/sec \)[/tex].
(c) Describe the slope of the secant line through [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex].
The slope of the secant line is the same as the average rate of change:
[tex]\[
\text{Slope of secant line} = 15 \, ft/sec
\][/tex]
Since the value is positive, the slope of the secant line is positive.
(d) Find the equation of the secant line through [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex]. Graph the secant line in the same viewing window as your position function.
To find the equation of the secant line, we use the point-slope form of the line equation [tex]\( y = m (t - t_1) + y_1 \)[/tex], where [tex]\( m \)[/tex] is the slope (15) and [tex]\((t_1, y_1)\)[/tex] is the point (0, 6):
[tex]\[
r(t) = 15 (t - 0) + 6 = 15 t + 6
\][/tex]
Therefore, the equation of the secant line is:
[tex]\[
r(t) = 15t + 6
\][/tex]
This completes the detailed, step-by-step solution for each part of the problem.
