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
To find the times when the bike is 15 feet away from the checkpoint, we can set the distance equation equal to 15 and solve for t:
|130 - 25t| = 15
Considering the two possible cases when the expression inside the absolute value is positive or negative:
130 - 25t = 15
-25t = -115
t = 4.6 seconds
-(130 - 25t) = 15
-130 + 25t = 15
25t = 145
t = 5.8 seconds
The bike is 15 feet away from the checkpoint at t = 4.6 seconds and t = 5.8 seconds.
Therefore, the answer is c. 4.6 sec and 5.8 sec.
To find the times at which the bike is 15 feet away from the checkpoint, we can set the equation d = |130 - 25t| equal to 15 and solve for t.
Given: d = |130 - 25t| = 15
We can split this equation into two cases, one where 130 - 25t is positive and one where it is negative. Let's consider each case separately:
Case 1: 130 - 25t > 0
In this case, the equation becomes 130 - 25t = 15.
Let's solve for t:
-25t + 130 = 15
-25t = 15 - 130
-25t = -115
t = -115 / -25
t = 4.6
So, in this case, t = 4.6 sec.
Case 2: 130 - 25t < 0
In this case, the equation becomes -(130 - 25t) = 15.
Let's solve for t:
130 - 25t = -15
-25t = -15 - 130
-25t = -145
t = -145 / -25
t = 5.8
So, in this case, t = 5.8 sec.
Therefore, the bike is 15 feet away from the checkpoint at t = 4.6 sec and t = 5.8 sec.
The correct answer is option c. 4.6 sec and 5.8 sec.
To determine the times when the bike is 15 feet away from the checkpoint, we need to find the values of t that solve the equation d = 15.
Given the equation d = |130 - 25t|, we can substitute 15 for d:
15 = |130 - 25t|
To eliminate the absolute value, we need to consider both positive and negative solutions:
130 - 25t = 15 ----> 115 = 25t ----> t = 4.6 sec
-(130 - 25t) = 15 ----> -130 + 25t = 15 ----> 25t = 145 ----> t = 5.8 sec
So, the bike is 15 feet away from the checkpoint at 4.6 seconds and 5.8 seconds.
Therefore, the correct answer is c. 4.6 sec and 5.8 sec.