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Chapter 1: Problem 68

(a) use the position equation \(s=-16 t^{2}+v_{0} t+s_{0}\) to write a function that represents the situation, (b) use a graphing utility to graph the function, (c) find the average rate of change of the function from \(t_{1}\) to \(t_{2},\) (d) describe the slope of the secant line through \(t_{1}\) and \(t_{2},\) (e) find the equation of the secant line through \(t_{1}\) and \(t_{2},\) and (f) graph the secant line in the same viewing window as your position function. An object is thrown upward from a height of 6.5 feet at a velocity of 72 feet per second. \(t_{1}=0, t_{2}=4\)

Short Answer

Expert verified
The position function is \(s(t) = -16t^2 + 72t + 6.5\), which represents the height of the object at time \(t\). The average rate of change from \(t_1=0\) to \(t_2=4\) is \(20\) feet per second. The slope of the secant line through \(t_1\) and \(t_2\) represents this average rate of change. The equation of this secant line is \(y = 20x + 6.5\).

Step by step solution

01

- Write the Position Function

Use the given position equation \(s=-16 t^{2}+v_{0} t+s_{0}\) to represent the specific situation by substituting given initial height (\(s_0=6.5\)) and initial velocity (\(v_0=72\)) into it. This yields the function \(s(t) = -16t^2 + 72t + 6.5\).
02

- Draw Function Graph

To graph the function \(s(t) = -16t^2 + 72t + 6.5\), use a graphing utility. The plot will show a downwards facing parabola representing the ascent, peak, and descent of the thrown object.
03

- Find Average Rate of Change

To find the average rate of change of the function from \(t_1\) to \(t_2\), use the formula \(\frac{f(t_2) - f(t_1)}{t_2 - t_1}\). Substituting \(t_1=0\) and \(t_2=4\), and using \(s(t)= -16t^2+72t+6.5\), we get \(\frac{s(4)-s(0)}{4-0}\). This equals \(\frac{-16*4^2 + 72*4 + 6.5 - 6.5}{4} = 20\) feet per second.
04

- Describe the Secant Line Slope

The slope of the secant line through \(t_1\) and \(t_2\) is the computed average rate of change, which is \(20\) feet per second. This slope signifies that for every second passed, the object descends by \(20\) feet in average between \(t=0\) and \(t=4\) seconds.
05

- Find the Secant Line Equation

The equation of the secant line can be found with the formula \(y - y_1 = m(x - x_1)\) where \(m\) is the slope and \((x_1, y_1)\) is a point on the line. Since \(t_1 = 0\), we can use the point \((0, s(0)) = (0, 6.5)\) and \(m = 20\). Thus, the equation is \(y - 6.5 = 20(x - 0)\), simplified into \(y = 20x + 6.5\)
06

- Draw the Secant Line

Draw the secant line with the equation \(y = 20x + 6.5\) in the same viewing window as the position function to visualize the average rate of change of object position over time. This is a straight line that intersects points \((t_i, s(t_i))\) for \(i=1,2\).

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