Starting from 130 feet away, a person on bicycle 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= |130-25t| At what times is the bike 15 feet away from the checkpoint?
To find out at what times the bike is 15 feet away from the checkpoint, we need to solve the equation |130-25t| = 15.
When |130-25t| = 15, it means that either 130-25t = 15 or 130-25t = -15.
So we have two equations to solve:
130-25t = 15 and 130-25t = -15.
For the first equation:
130-25t = 15.
Subtract 130 from both sides of the equation:
-25t = 15 - 130
-25t = -115
Divide both sides by -25 (remember that dividing by a negative number flips the inequality sign):
t = -115 / -25
t = 4.6
For the second equation:
130-25t = -15.
Subtract 130 from both sides of the equation:
-25t = -15 - 130
-25t = -145
Divide both sides by -25:
t = -145 / -25
t = 5.8
So the bike is 15 feet away from the checkpoint at times t = 4.6 seconds and t = 5.8 seconds.