SOLUTION: Two cars are approaching an intersection. One is 3 miles South of the intersection and is moving at a constant speed of 40 mph. At the same time, the other car is 4 miles East of t

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Question 339836: Two cars are approaching an intersection. One is 3 miles South of the intersection and is moving at a constant speed of 40 mph. At the same time, the other car is 4 miles East of the intersection and is moving at a constant speed of 50 mph. Express the distance d between the cars as a function of time t.
Found 2 solutions by mananth, jrfrunner:
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
one car 3 miles 40 mph ----- t1 = 3/40 hours
second car 4 miles 50 mph----t2= 4/50 hours
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they form a right angle with intersection.
the distance between the cars = hypotenuse.
..

Take square root of both sides

Answer by jrfrunner(365) (Show Source):
You can put this solution on YOUR website!
you basically have a right triangle formed by the movement of the cars, one from coming from the east and one coming from the south
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let d=ditance between the cars.
let x=distance from car from the east to the meeting point
let y=distance from car from the south to the meeting point
distance=speed*time; d=s*t
let =Distance by car from east
let =Distance by car from south
let ="shrinking" distance by car from east
let ="shrinking" distance by car from south
let =Distance between cars
let ="shrinking" distance between cars
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at t=0, one car is 4 miles east and the other car is 3 miles south, so the distance between them, by the pythagorean theorem:

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as time goes forward, the distances from the east and the south shrink toward their meeting point.
car from the east travels a distance: in general
relative to the meeting point 4 miles away the distance shrinks
as
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car from the south travels a distance: in general
relative to the meeting point 3 miles away the distance shrinks
as
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