Question

1. a ball is thrown straight upward. the velocity of the ball 2 seconds after being thrown is 52 feet per second. the velocity after 6 seconds is 12 feet per second. assume the velocity changes at a constant rate.

part a:
calculate the slope of the line that models the velocity.
______ ft/sec

part b:
write the equation of the line in point - slope form.
y - ____ = __(x - ____)

Explanation:

PART A:

Step1: Identify two points

Let \( x \) be time in seconds and \( y \) be velocity in ft/sec. The two points are \( (2, 52) \) and \( (6, 12) \).

Step2: Use slope formula

The slope \( m \) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Substituting \( (x_1,y_1)=(2,52) \) and \( (x_2,y_2)=(6,12) \), we get \( m=\frac{12 - 52}{6 - 2}=\frac{-40}{4}=- 10 \).

Step1: Recall point - slope form

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( m \) is the slope and \( (x_1,y_1) \) is a point on the line. We know \( m=-10 \), and we can use the point \( (2,52) \) (we could also use \( (6,12) \)).

Step2: Substitute into point - slope form

Substituting \( m = - 10 \), \( x_1=2 \), and \( y_1 = 52 \) into the point - slope formula \( y - y_1=m(x - x_1) \), we get \( y - 52=-10(x - 2) \).

Answer:

\(-10\)

PART B: