SOLUTION: A family left your house in 2 cars at the same time 1 car travel an avg of 7 miles per hour faster than the other. When the first arrive at the destination after 5.5 hours of dri

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Question 953674:
A family left your house in 2 cars at the same time 1 car travel an avg of 7 miles per hour faster than the other. When the first arrive at the destination after 5.5 hours of driving, cars had driven a total of 599.5 miles if this second car continues the same average speed, how much time, to the nearest minute, will it take before the second car arrives?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the speed of the slower car = +s+ mi/hr
The speed of the faster car = +s+%2B+7+ mi/hr
-----------------
The 1st to arrive at destination is the faster car
Equation for faster car:
(1) +d+=+%28+s+%2B+7+%29%2A5.5+
Equation for slower car:
(2) +599.5+-+d+=+s%2A5.5+
-------------------------
(1) +d+=+5.5s+%2B+38.5+
(1) +d+-+5.5s+=+38.5+
and
(2) +d+%2B+5.5s+=+599.5+
-------------------------
Add the equations
+2d+=+638+
+d+=+319+
The slower car has traveled
+599.5+-+319+=+280.5+
The slower car has +319+-+280.5+=+38.5+ mi
left to reach destination
-----------------------
Equation for slower car:
(2) +599.5+-+d+=+s%2A5.5+
(2) +599.5+-+319+=+s%2A5.5+
(2) +280.5+=+5.5s+
(2) +s+=+51+
-----------------
Find time for slower car to reach destination:
+38.5+=+51t+
+t+=+.755+ hrs
convert to minutes:
+t+=+45.29+ min
To the nearest minutes, it takes 45 min
for slower car to arrive
----------------------
check:
For slower car:
+.755+%2B+5.5+=+6.255+
+d%5B1%5D+=+319+
+d%5B1%5D+=+s%2A6.255+
+319+=+6.255s+
+s+=+50.999+
This is close enough to +s+=+51+
OK