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?

Answer 1

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.