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?

Question

(1 point)
Responses

4.6 sec and 9.2 sec

2.9 sec and 5.8 sec

4.6 sec and 5.8 sec

2.9 sec and 3.3 sec

Answer 1

The distance between the bicycle and the checkpoint is given by the equation d = 130 - 25t, where d is the distance and t is the time in seconds.

We need to find the times when the bike is 15 feet away from the checkpoint. So we set d = 15 and solve for t:

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

Thus, the bike is 15 feet away from the checkpoint at 4.6 seconds.

Now, let's check the other possible solutions:

2.9 seconds:
d = 130 - 25*2.9
d = 130 - 72.5
d ≈ 57.5 feet (not 15 feet)

3.3 seconds:
d = 130 - 25*3.3
d = 130 - 82.5
d ≈ 47.5 feet (not 15 feet)

5.8 seconds:
d = 130 - 25*5.8
d = 130 - 145
d = -15 feet (not a valid distance)

So, the only valid solution is t = 4.6 seconds. The bike is 15 feet away from the checkpoint at 4.6 seconds. The correct option is therefore:

4.6 sec and 9.2 sec.