Start from 110 feet away, a person on a bicycle rides towards a checkpoint and then passes it. The rider is travelling at a constant rate of 30 feet per second. The distance between the bicycle and the checkpoint is given by the equation d=|110-30t| at what times is the bike 40 feet away from the checkpoint?
a. 2.3 sec and 5 sec
b. 1.6 sec and 5 sec
c. 2.3 sec and 4.6 sec
d. 1.6 sec and 4.6 sec
|110-30t| = 40
so, either
110-30t = 40
30t = 70
t = 7/3 = 2.3
or
-(110-30t) = 40
-110+30t = 40
30t = 150
t = 5
So, (A) is correct
Thank you so much!
Steve definitely got it right!
Just did the test still correct in 2020!!!
Thanks Steve!
To find the times when the bike is 40 feet away from the checkpoint, we need to solve the equation: d = |110 - 30t| = 40.
First, let's express the equation with its two possible forms:
1. When the distance is positive:
110 - 30t = 40
2. When the distance is negative, we take the absolute value of the equation:
-(110 - 30t) = 40
Now, let's solve both equations:
1. When the distance is positive:
110 - 30t = 40
Subtract 110 from both sides:
-30t = -70
Divide both sides by -30 (since we want to isolate t):
t = 70/30
t = 7/3 or approximately 2.3 seconds
2. When the distance is negative (taking the absolute value):
-(110 - 30t) = 40
Distribute the negative sign:
-110 + 30t = 40
Add 110 to both sides:
30t = 150
Divide both sides by 30:
t = 150/30
t = 5 seconds
So the bike is 40 feet away from the checkpoint at 2.3 seconds and 5 seconds.
The correct answer is option a. 2.3 sec and 5 sec.