Question

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. At what times is the bike 15 feet away from the checkpoint?

Answer 1

is there another amount of seconds?

Answer 2

so would 4.6 sec and 5.8 sec be correct?

Answer 3

Let's denote the time in seconds as t.
The rider is traveling at a constant rate of 25 feet per second, so the distance between the bicycle and the checkpoint is given by the equation:
distance = 130 - (25t)
To find the times when the bike is 15 feet away from the checkpoint, we set the distance equal to 15 feet:
130 - 25t = 15
Solving for t:
25t = 130 - 15
25t = 115
t = 115/25
t = 4.6 seconds
Therefore, the bike is 15 feet away from the checkpoint at t = 4.6 seconds.

Answer 4

Apologies for the confusion.
To find any other time when the bike is 15 feet away from the checkpoint, we can substitute the distance equation:

130 - 25t = 15

Simplifying the equation:

25t = 130 - 15
25t = 115
t = 115/25
t = 4.6 seconds

So, the bike is 15 feet away from the checkpoint at t = 4.6 seconds. There is no other time when the bike is exactly 15 feet away from the checkpoint.