SOLUTION: a bicyclist leaves a point and then a car leaves the same point 2 hours later going 12mph faster. After 90 miles the car catches the bicyclist. Find the speed of both the car and t
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Question 426723: a bicyclist leaves a point and then a car leaves the same point 2 hours later going 12mph faster. After 90 miles the car catches the bicyclist. Find the speed of both the car and the bicyclist.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
a bicyclist leaves a point and then a car leaves the same point 2 hours later going 12mph faster.
After 90 miles the car catches the bicyclist.
Find the speed of both the car and the bicyclist.
:
Let s = the bike speed
then
(s+12) = the car speed
:
Write a time equation: time = dist/speed
:
Bike time = Car time + 2 hrs
= + 2
:
Multiply by s(s+12), results
90(s+12) = 90s + 2s(s+12)
:
90s + 1080 = 90s + 2s^2 + 24s
:
Arrange as a quadratic equation on the right
0 = 90s - 90s + 2s^2 + 24s - 1080
:
2s^2 + 24s - 1080 = 0
:
Simplify,divide by 2
s^2 + 12s - 540 = 0
Factors to
(s-18)(s+30) = 0
the positive solution
s = 18 mph is the bike speed
then
18 + 12 = 30 mph is the car's
:
:
Check solution by finding the actual travel time of each
90/18 = 5 hrs
90/30 = 3 hrs
--------------
a diff of 2 hrs
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