Starting from 150
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 dequals
StartAbsoluteValue 150 minus 20 t EndAbsoluteValue.
At what times is she 50
ft from you?
Question content area bottom
Part 1
She is 50
ft away from you after enter your response here
s.
(Use a comma to separate answers as needed.)
To find the times at which she is 50 ft away from you, we can set up the equation:
|150 - 20t| = 50
First, let's consider the case when 150 - 20t > 0:
150 - 20t = 50
-20t = -100
t = 5
When t = 5 seconds, she is 50 ft away from you.
Now let's consider the case when 150 - 20t < 0:
-(150 - 20t) = 50
-150 + 20t = 50
20t = 200
t = 10
When t = 10 seconds, she is also 50 ft away from you.
Therefore, she is 50 ft away from you at t = 5 seconds and t = 10 seconds.