SOLUTION: I've done this problem 5 times and I can't figure it out. Please help... The winner of the track meet had an average speed of 5 meters per second. The second place runner had an a

Question 831853: I've done this problem 5 times and I can't figure it out. Please help...
The winner of the track meet had an average speed of 5 meters per second. The second place runner had an average speed of 4.5 meters per second. If the winner finished 2.2 seconds ahead of the second place runner, how long did it take the winner to cross the finish line?
Using this equation:
Rate of Winner x Time of Winner (t) = Rate of 2nd Place Runner x Time of 2nd Place Runner (t+2.2)
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
---
s = d/t
t = d/s
---
x = race distance
y = time of winner
---
time equation for winner: using t = d/s
y = x/5
---
time equation for second place runner: using t = d/s
y + 2.2 = x/4.5
y = x/4.5 - 2.2
---
linear system:
y = x/5
y = x/4.5 - 2.2
---
y = y
so the right-hand sides of the two equations above, are equal to each other:
---
x/5 = x/4.5 - 2.2
x/4.5 - x/5 = 2.2
5x/5*4.5 - 4.5x/5*4.5 = 2.2
5x - 4.5x = 2.2*5*4.5
0.5x = 2.2*5*4.5
x = 2.2*5*4.5/0.5
x = 99 m
---
y = x/5
y = 99/5
y = 19.8
---
answer:
winner's time:
y = 19.8 seconds
---
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