Exercise 63 Page 212

To identify the various parts of this equation, we need to first recognize that this equation is an adaptation of the standard Distance Formula, d=r* t where d is distance, r is the rate, and t is the time. With this in mind, we can recognize that 18 must be our rate, which is measured in feet per second. But, inside the absolute value we have: 200-18t.We can know that 200 is the initial distance away from you because it is constant in an equation about distance. We can solve for the time(s) when your friend is 120 ft away from you by substituting d= 120 into our equation. d=|200-18t| ⇒ 120=|200-18t| When solving an absolute value equation we need to remember to look at both the positive and negative cases. &Case $1$: 120=200-18t &Case $2$: -120=200-18t Now, we can solve these cases separately. Let's start with the positive case and convert any improper fractions into mixed numbers.

Our first solution is that your friend will be 120 feet away from you when 4 49 seconds have passed. Let's take a look at the negative case.

Our second solution is that your friend will be 120 feet away from you again when 17 79 seconds have passed.