SOLUTION: Kiran drove from Tortula to Cactus, a distance of 212 mi. She increased her speed by 13 mi/h for the 385 mi trip from Cactus to Dry Junction. If the total trip took 10 h, what was
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-> SOLUTION: Kiran drove from Tortula to Cactus, a distance of 212 mi. She increased her speed by 13 mi/h for the 385 mi trip from Cactus to Dry Junction. If the total trip took 10 h, what was
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Question 990904: Kiran drove from Tortula to Cactus, a distance of 212 mi. She increased her speed by 13 mi/h for the 385 mi trip from Cactus to Dry Junction. If the total trip took 10 h, what was her speed from Tortula to Cactus? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Kiran drove from Tortula to Cactus, a distance of 212 mi.
She increased her speed by 13 mi/h for the 385 mi trip from Cactus to Dry Junction.
If the total trip took 10 h, what was her speed from Tortula to Cactus?
:
let s = speed from T to C
then
(s+13) = speed from C to D
:
Write a time equation, time = dist/speed + = 10 hrs
multiply equation by s(s+13)
s(s+13)* + s(s+13)* = 10s(s+13)
Cancel the denominators
212(s+13) + 385s = 10s^2 + 130s
212s + 2756 + 385s = 10s^2 + 130s
597s + 2756 = 10s^2 + 130s
Arrange as a quadratic equation
10s^2 + 130s - 597s - 2756 = 0
10s^2 - 467s - 2756 = 0
Use the quadratic formula to find s: a=10; b=-467; c=-2756
I got a positive solution of 52 mph from T to C
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you can check this for yourself + =
