SOLUTION: a car leaves san diego at 1 pm traveling at a constant rate of 40 mph toward los angeles. thirty minutes later, another car leaves san diego traveling toward los angeles at a const

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Question 342281: a car leaves san diego at 1 pm traveling at a constant rate of 40 mph toward los angeles. thirty minutes later, another car leaves san diego traveling toward los angeles at a constant speed of 55 mph. after what time is the distance between the cars 15 miles?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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a car leaves san diego at 1 pm traveling at a constant rate of 40 mph toward los angeles.
thirty minutes later, another car leaves san diego traveling toward los angeles
at a constant speed of 55 mph.
after what time is the distance between the cars 15 miles?
:
Let t = travel time of the faster car
then
(t+.5) = travel time of the 1st car
:
Write a distance equation, dist = speed * time
:
slow car dist - fast car dist = 15 miles
40(t+.5) - 55t = 15 mi
40t + 20 - 55t = 15
40t - 55t = 15 - 20
-15t = -5
t = %28-5%29%2F%28-15%29
t = + 1%2F3 hr travel time of the faster car
that's 20 min and that car left a 1:30
therefore
1:50 is the time when the cars will be 15 mi apart
:
:
Confirm this by finding the distance traveled by each car (20 min ~.33hr)
40(.5+.33) = 33.2 mi
55(.33) = 18.2 mi
---------------------
difference: 15 mi