SOLUTION: please help me solve:
Kevin drove 320 miles to a resort. He return trip took 20 min longer because his speed returning was 4 mi/h slower than going
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Kevin drove 320 miles to a resort. He return trip took 20 min longer because his speed returning was 4 mi/h slower than going
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Question 784848: please help me solve:
Kevin drove 320 miles to a resort. He return trip took 20 min longer because his speed returning was 4 mi/h slower than going Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Kevin drove 320 miles to a resort.
He return trip took 20 min longer because his speed returning was 4 mi/h slower than going
:
Let (s+4) = speed of the outbound trip
then
s = speed of the return trip
:
Change 20 min to 1/3 hr
:
Write a time equation; time = dist/speed
:
return time - outbound time = 20 min - =
multiply by 3s(s+4) to clear the denominators, resulting in:\
3(s+4)(320) - 3s(320) = s(s+4)
960s + 3840 - 960s = s^2 + 4s
:
Arrange as a quadratic equation on the right
0 = s^2 + 4s - 3840
:
You can use the quadratic formula here but this will factor to
(s+64)(s-60) = 0
the positive solution
s = 60 mph on the return trip
and obviously 64 mph on the outbound trip
:
:
Confirm this by finding the actual times
320/60 = 5.33
320/64 = 5.00
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dif: .33 hrs which is 20 min
