Starting from 130 feet away, a person on a bicycle rides towards a checkpoint and then passes it. The rider is traveling as 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.
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
It's c
Thanks Nunya!
how long does it take to go the 130 ft?
130/25 = 5.2 sec
How long to go 15 ft?
0.6 sec
So, at 5.2±0.6 the distance will be 15 ft
Now what do you think?
is it a or c i don't understand what the second part means 5.8 seconds or 9.2 seconds?
To find the times when the bike is 15 feet away from the checkpoint, we need to set up the equation:
d = |130 - 25t| = 15
To solve this equation, we need to consider two scenarios:
1. When the expression inside the absolute value is positive:
130 - 25t = 15
Solving for t in this case gives us:
-25t = 15 - 130
-25t = -115
t = (-115)/(-25)
t = 4.6 sec
2. When the expression inside the absolute value is negative:
130 - 25t = -15
Solving for t in this case gives us:
-25t = -15 - 130
-25t = -145
t = (-145)/(-25)
t = 5.8 sec
So, the correct option is c. 4.6 sec and 5.8 sec.