Question

Start from 110 feet away, a person on a bicycle rides towards a checkpoint and then passes it. The rider is traveling at a constant rate of 30 feet per second. The distance between the bicycle and the checkpoint is given by the equation d equals absolute value of the quantity 110 minus 30 t. 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

Answer 1

The bicycle is 40 feet away from the checkpoint at 2.3 seconds and 5 seconds, calculated by setting the absolute value distance equation equal to 40 and solving for time t.

Step-by-step explanation:

The student wants to know at what times the bike is 40 feet away from the checkpoint when the bicycle starts 110 feet away and moves toward the checkpoint at a constant rate of 30 feet per second. The distance from the checkpoint at any time t, in seconds, is represented by the equation d = |110 - 30t|. To find the times when the bike is 40 feet away, we set this equation equal to 40 and solve for t.

Let's solve the equation: |110 - 30t| = 40.

We consider two cases because the absolute value of a number can be both positive and negative:

1. 110 - 30t = 40: Solving for t gives t = 70 / 30 = 2.3 seconds.
2. 110 - 30t = -40: Solving for t gives t = 150 / 30 = 5 seconds.

Therefore, the bike is 40 feet away from the checkpoint at 2.3 seconds and 5 seconds.