Starting from 100 ft away, your friend skates toward you and then passes you. She skates at a constant speed of 20 ft's. Her distance d from you in feet after t seconds is given by d = |100 - 20t| At what times is she 80 from you?
She is 80 ft away from you __ s .
(Use a comma to separate answers as needed.)
To find the time when she is 80 ft away from you, we can set up the equation:
|100 - 20t| = 80
Let's solve for t in this equation.
When 100 - 20t is positive:
100 - 20t = 80
-20t = -20
t = 1
When 100 - 20t is negative:
-(100 - 20t) = 80
-100 + 20t = 80
20t = 180
t = 9
Therefore, she is 80 ft away from you at t = 1 s and t = 9 s.
She is 80 ft away from you at 1 s, 9 s.
To find the times when she is 80 ft away from you, we can set up the equation:
|100 - 20t| = 80
To solve this equation, we can split it into two separate cases:
Case 1: 100 - 20t = 80
Solving for t in this case:
100 - 20t = 80
-20t = 80 - 100
-20t = -20
t = (-20) / (-20)
t = 1
Case 2: 100 - 20t = -80
Solving for t in this case:
100 - 20t = -80
-20t = -80 - 100
-20t = -180
t = (-180) / (-20)
t = 9
Therefore, she is 80 ft away from you at t = 1 s and t = 9 s.
To find the times when she is 80 ft away from you, we can substitute 80 for d in the equation and solve for t.
Given the distance equation: d = |100 - 20t|
Substituting 80 for d, we have: 80 = |100 - 20t|
Now, let's solve for t by isolating it:
Case 1: 100 - 20t is positive (inside the absolute value)
80 = 100 - 20t
-20t = -20
t = 1
Case 2: 100 - 20t is negative (inside the absolute value)
80 = -(100 - 20t)
80 = -100 + 20t
20t = 180
t = 9
So, she is 80 ft away from you at t = 1 second and t = 9 seconds.