HELP NEEDED WITH ABSOLUTE VALUE INEQUALITIES WORD PROBLEMS

Question

A car is 500 ft from Krista and driving toward her at 32 ft per second. At what times will the car be less than 50 ft away from her? Write an appropriate absolute value inequality for the given situation and solve.

Answer 1

So there are two ways to approach this problem. First of all, let's use logic! If a car is 500 feet away, then it has to travel either almost all the way to Krista (that is, 450 of the 500 feet) OR 50 feet PAST Krista to be withion 50 feet of her.
So now we know that if the car has gone between 450 and 550 feet, it is within 50 feet of Krista.

Next we have to think, "well how do we measure DISTANCE?"

We use: D = r * t, so the rate of 32 times the time in seconds will equal the distance traveled by the car. We can express this distance as "32t." We want 32t to be between 450 and 550 feet. So let's take the AVERAGE of those 2 distances: That will be the SUM of 450 & 550 divided by 2, which is 500. Now let's take the averages of the DIFFERENCE between the 2 distances: 550-450=100, divided by 2 = 50.

So we started out logically by thinking about the distance of 50 feet away from Krista and got that the distance must be between these two points:
450 > D > 550
To write this as an absolute value inequality, we would do:
|32t - 500| < 50 which would break out into TWO problems:
32t-500 <(+50) and 32t-500>(-50)
(Remember on the second one to FLIP the sign!)
Solving each for t gives us, roughly:
14.06 < t < 17.18
So after 10 seconds, for example, the car would only have gone 320 feet, but after 15 seconds the car will have gone 480 feet which is 20 feet away from Krista, and after 20 seconds the car will be at 640 feet, and thus, far away from Krista now.

Make sense? Notice when you have a word problems requiring you to figure out a measurement BETWEEN 2 points, first we add the 2 numbers and take the average, and then we SUBTRACT the 2 numbers and take that average. That gives us the absolute value inequality.

Measure must be between LOW and HIGH number.
Looks like this:
|M-(LOW+HIGH)/2|<(HIGH-LOW)/2
Example, if a cookie has to be between 5 and 7 inches to be considered standard, then it would look like this:
|C-6|<1 (where cookie = C)

Hope this helps!