SOLUTION: Sarah travels for 100 miles and then decreases her rate of travel by 5 mph and continued the remaining 300 miles of her trip. If the total trip took 7 hours and 40 minutes, at what
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: Sarah travels for 100 miles and then decreases her rate of travel by 5 mph and continued the remaining 300 miles of her trip. If the total trip took 7 hours and 40 minutes, at what
Log On
Question 976017: Sarah travels for 100 miles and then decreases her rate of travel by 5 mph and continued the remaining 300 miles of her trip. If the total trip took 7 hours and 40 minutes, at what two speeds did she travel? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Sarah travels for 100 miles and then decreases her rate of travel by 5 mph and continued the remaining 300 miles of her trip.
If the total trip took 7 hours and 40 minutes, at what two speeds did she travel?
:
Let s = the original speed for the 1st 100 mi
then
(s-5) = the speed for the remaining 300 mi
:
Change 7 hrs 40 min to hrs: 7 + 40/60 = 7 =
:
Write a time equation; time = dist/speed
: + =
multiply by 3s(s-5), cancel the denominators and you have
3(s-5)*100 + 3s(300) = 23s(s-5)
300(s-5) + 3s(300) = 23s^2 - 115s
300s - 1500 + 900s = 23s^2 - 115s
Combine like terms on the right
0 = 23s^2 - 115s - 1200s + 1500
23s^2 - 1315s + 1500 = 0
use the quadratic formula to find s: a=23; b=-1315; c=1500
I got the reasonable solution of
s = 56 mph the 1st 100 mi
therefore
51 mph the last 300 mi (5 mph slower)
:
:
Confirm this by finding the actual time at each speed
100/56 = 1.79 hrs
300/51 = 5.88 hrs
-----------------
total time: 7.67 hrs which is about 7 hrs 40 min
