Tricia and janine are roommates and leave Houston on Interstate 10 at the same time to visit their families for a long weekend. Tricia travels west and Janine travels east. If Tricia's average speed is 12 mph faster than Janine's, find the speed of each if they are 320miles apart in 20 hours and 30 minutes.
Janine's speed --- x
Tricia's speed --- x+12
Janince's distance = 20.5(x)
Tricia's distance = 20.5(x+12)
20.5x + 20.5(x+12) = 320
41x + 246 = 320
41x = 74
x = 1.8
So Janine went at 1.8 mph and Tricia went at 13.8 mph ??????
The answers do check out mathematically but don't make a lot of sense.
To solve this problem, we can set up a system of equations based on the given information. Let's denote the speed of Janine as "x" mph. Since Tricia's average speed is 12 mph faster, we can represent her speed as "x + 12" mph.
We know that distance equals speed multiplied by time. Therefore, we can set up the following equations:
Tricia's distance: Distance = Speed * Time
Janine's distance: Distance = Speed * Time
For Tricia:
Distance = (x + 12) * Time
For Janine:
Distance = x * Time
Given that they are 320 miles apart, we have the equation:
Distance + Distance = 320 miles
Substituting the distances and using the given time of 20 hours and 30 minutes (which is equivalent to 20.5 hours), we can set up the equation:
(x + 12) * 20.5 + x * 20.5 = 320
Now, let's solve for x to find Janine's speed:
20.5x + 246 + 20.5x = 320
41x + 246 = 320
41x = 320 - 246
41x = 74
x = 74 / 41
x ≈ 1.804 mph (rounded to three decimal places)
So, Janine's speed is approximately 1.804 mph.
To find Tricia's speed, we can substitute this value back into the equation:
Tricia's speed = x + 12
≈ 1.804 + 12
≈ 13.804 mph (rounded to three decimal places)
Therefore, Janine's speed is approximately 1.804 mph, and Tricia's speed is approximately 13.804 mph.