Question

consider the position function ( s(t) = -16t^2 + 120t ) representing the position of an object moving vertically along a line. sketch a graph of ( s ) with the secant line passing through... determine the slope of the secant line and explain its relationship to the moving object.

Explanation:

To determine the slope of the secant line for the position function \( s(t) = -16t^2 + 120t \) (assuming the points are \( t = 0 \) and \( t = 5 \), as it's a common setup for such problems):

Step 1: Find \( s(0) \)

Substitute \( t = 0 \) into \( s(t) \):
\( s(0) = -16(0)^2 + 120(0) = 0 \).

Step 2: Find \( s(5) \)

Substitute \( t = 5 \) into \( s(t) \):
\( s(5) = -16(5)^2 + 120(5) = -16(25) + 600 = -400 + 600 = 200 \).

Step 3: Calculate the slope of the secant line

The slope \( m \) of the secant line through \( (t_1, s(t_1)) \) and \( (t_2, s(t_2)) \) is given by:
\( m = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \).

Using \( t_1 = 0 \), \( s(t_1) = 0 \), \( t_2 = 5 \), \( s(t_2) = 200 \):
\( m = \frac{200 - 0}{5 - 0} = \frac{200}{5} = 40 \).

The slope of the secant line is \( 40 \). This slope represents the average velocity of the object between \( t = 0 \) and \( t = 5 \) seconds (since velocity is the rate of change of position over time). A positive slope means the object’s average velocity is upward (positive direction) during this interval.

To sketch the graph:

• The position function \( s(t) = -16t^2 + 120t \) is a downward - opening parabola (since the coefficient of \( t^2 \) is negative).
• Plot the points \( (0, 0) \) and \( (5, 200) \).
• Draw a straight line (the secant line) connecting these two points. The slope of this line is \( 40 \), so for every 1 - unit increase in \( t \), \( s(t) \) increases by 40 units on average between \( t = 0 \) and \( t = 5 \).

Answer:

The slope of the secant line is \( \boldsymbol{40} \).