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? (1 point) Responses
A. 4.6 sec and 9.2 sec
B. 2.9 sec and 5.8 sec
C. 4.6 sec and 5.8 sec
D.2.9 sec and 3.3 sec
To find the times when the bike is 15 feet away from the checkpoint, we can set up an equation using the given information.
The distance between the bike and the checkpoint at any given time (t) can be represented by the equation:
Distance = Initial distance - rate x time
In this case, the initial distance is 130 feet, the rate is 25 feet per second, and the desired distance is 15 feet.
So, the equation becomes:
15 = 130 - 25t
Rearranging the equation:
25t = 130 - 15
25t = 115
t = 115/25
t = 4.6
Therefore, the bike is 15 feet away from the checkpoint at 4.6 seconds.
To find another time when the bike is 15 feet away, we can substitute 15 for the distance in the equation and solve for t:
15 = 130 - 25t
25t = 130 - 15
25t = 115
t = 115/25
t = 4.6
Therefore, the bike is also 15 feet away from the checkpoint at 4.6 seconds.
Thus, the answer is A. 4.6 sec and 9.2 sec.