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 he checkpoint is given by the equation d = |110 - 30t|. At what time is the bike 40 feet away from the checkpoint?
2.3 sec and 5 sec!
does anyone know the answers to the whole test?
To find the time at which the bike is 40 feet away from the checkpoint, we can set up the equation d = 40 and solve for t.
The given equation for the distance is d = |110 - 30t|. Substituting d with 40, we get:
40 = |110 - 30t|
Now, we can solve for t by considering the two cases when the expression inside the absolute value is positive and negative.
Case 1: (110 - 30t) is positive
40 = 110 - 30t
Rearranging the equation:
30t = 110 - 40
30t = 70
Dividing both sides by 30:
t = 70 / 30
t = 7/3
Case 2: (110 - 30t) is negative
40 = -(110 - 30t)
Simplifying the equation by applying the negative sign inside the parentheses:
40 = -110 + 30t
Rearranging the equation:
30t = 40 + 110
30t = 150
Dividing both sides by 30:
t = 150 / 30
t = 5
Since we have two possible values for t, we need to determine which one satisfies the original equation.
When t = 7/3, substituting it into the original equation:
d = |110 - 30t|
d = |110 - 30(7/3)|
d = |110 - 70|
d = |40|
The distance is equal to 40, which means this value of t is correct.
Therefore, the bike is 40 feet away from the checkpoint at t = 7/3 seconds.