starting from 1.5 miles away, a car drives toward a speed checkpoint and then passes it. The car travels at a constant rate of 53 miles per hour. The distance of the car from the checkpoint is give by d=|1.5-53t|. At what time if the car o.1 miles from the checkpoint? Answer has to be in seconds.
No I put the question mark there because I didn't know if it was right or not.
So no i wasn't guessing lol
A. 95.s and 108.7s
B. 10.2s and 101.9s
C. 108.7s and 10.2s
D. 95.1s and 10.2s
Answer: A?
Suppose y varies directly with x, and y = 8 when x = -6. What direct variation equation relates x and y? What is the value of y when x = -2?
Oh, I see the car is racing towards the checkpoint! Let's put on our clown hat and solve this equation with a dash of humor.
Given the equation d = |1.5 - 53t| to find the time when the car is 0.1 miles away from the checkpoint, we need to set up the equation:
0.1 = |1.5 - 53t|
Now, let's split this equation into two cases because of the absolute value:
Case 1: When 1.5 - 53t is positive:
0.1 = 1.5 - 53t
Solving for t, we get:
-52.9 = -53t
t = (-52.9) / (-53)
t ≈ 0.996 seconds
Case 2: When 1.5 - 53t is negative:
0.1 = -(1.5 - 53t)
Solving for t, we get:
-0.9 = -1.5 + 53t
52.1 = 53t
t = 52.1 / 53
t ≈ 0.983 seconds
So, the car will be 0.1 miles away from the checkpoint approximately 0.996 seconds and 0.983 seconds after it starts its journey.
To find the time when the car is 0.1 miles from the checkpoint, we can set up the equation:
d = |1.5 - 53t| = 0.1
To solve this equation, we need to consider two cases: when the expression inside the absolute value is positive and when it is negative.
Case 1: 1.5 - 53t > 0
In this case, we can simplify the equation to:
1.5 - 53t = 0.1
Now, solve for t:
53t = 1.5 - 0.1
53t = 1.4
t = 1.4 / 53
t ≈ 0.0264 hours
Since we want the answer in seconds, we convert hours to seconds:
0.0264 hours × 60 minutes/hour × 60 seconds/minute ≈ 95.04 seconds
Case 2: 1.5 - 53t < 0
In this case, we flip the sign inside the absolute value, and our equation becomes:
-(1.5 - 53t) = 0.1
Simplify and solve for t:
-1.5 + 53t = 0.1
53t = 1.5 + 0.1
53t = 1.6
t = 1.6 / 53
t ≈ 0.0302 hours
Convert hours to seconds:
0.0302 hours × 60 minutes/hour × 60 seconds/minute ≈ 108.72 seconds
Thus, the car is 0.1 miles away from the checkpoint approximately at 95.04 seconds or 108.72 seconds after it started from 1.5 miles away.