SOLUTION: A car leaves Oak Corners at 11:30 AM traveling south at 60 mph. At the same time, another car is 50 miles west of Oak Corners traveling east at 40 mph. Express the distance d betwe

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Question 214875: A car leaves Oak Corners at 11:30 AM traveling south at 60 mph. At the same time, another car is 50 miles west of Oak Corners traveling east at 40 mph. Express the distance d between the cars as a function of the time t after the first car left Oak Corners.
Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
A car leaves Oak Corners at 11:30 AM traveling south at 60 mph. At the same time, another car is 50 miles west of Oak Corners traveling east at 40 mph. Express the distance d between the cars as a function of the time t after the first car left Oak Corners.
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It's not clear which is the "1st car", but starting from the time one car is 50 miles west (1130):
d^2 = (50 - 40t)^2 + (60t)^2
d^2 = 2500 - 4000t + 1600t^2 + 3600t^2
d^2 = 5200t^2 - 4000t + 2500
d(t) = 10sqrt(52t^2 - 40t + 25) t in hours, d in miles

Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
The car going south has mi From Oak Corners
where is in hours and is in miles
The car going east is mi from Oak Corners
The distance between the cars is:

answer
check:
In 3 hrs, the car going south has gone:

mi
In 3 hrs the car going east is

mi from Oak Corners (It's East of it)

mi
and

OK