Starting from 130 feet away, a person on bicycles rides towards a checkpoint and then passes it. The rider is traveling at a constant rate of 25 feet per second. The distance between the bicycle and the checkpoint is given by the equation d=|130-25d|. At what times is the bike 15 feet away from the checkpoint?
To find the times when the bike is 15 feet away from the checkpoint, we need to solve the equation:
d = |130 - 25t| = 15
First, let's consider the case when 130 - 25t is positive:
130 - 25t = 15
-25t = -115
t = 4.6
In this case, when t = 4.6, the bike is 15 feet away from the checkpoint.
Now, let's consider the case when 130 - 25t is negative:
-(130 - 25t) = 15
-130 + 25t = 15
25t = 145
t = 5.8
In this case, when t = 5.8, the bike is also 15 feet away from the checkpoint.
So, the bike is 15 feet away from the checkpoint at times t = 4.6 and t = 5.8 seconds.