Question 875439: The question is: Rachel drove 600 miles to her destination. The return trip was made at a speed that was 10 mph slower. If the total travel time was 22 hours, how fast did Rachel travel on each part of the trip?
So far I know that I have to make a table chart where I make two equations. The First trip and another with the return trip. I know that the distance formula is d=r*t. My distance is 600 miles and one rate is r and the other one is r-10, since it was going 10 mph slower, but I have no idea on what to put for time because it gives me a total not each time for the two trips.
Please help:(
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Trip to destination:
d = r*t
600 = r*t
600/r = t
t = 600/r
It takes 600/r hours to travel to the destination.
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Return Trip:
d = r*t
d = (r-10)*t
600 = (r-10)*t
600/(r-10) = t
t = 600/(r-10)
It takes 600/(r-10) hours to get back home.
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Add up the two times and set that equal to the total time 22
Time1 + Time2 = Total Time
600/r + 600/(r-10) = 22
Now we solve for r
600/r + 600/(r-10) = 22
r(r-10) = r(r-10)*22 ... multiply both sides by the LCD r(r-10) to clear out the fractions.
r(r-10) + r(r-10) = r(r-10)*22
(r-10)*600 + r*600 = r(r-10)*22 ... Notice how the fractions are gone after multiplying everything by the LCD
600(r-10) + 600r = 22r(r-10)
600r-6000 + 600r = 22r^2 - 220r
1200r-6000 = 22r^2 - 220r
0 = 22r^2 - 220r - 1200r + 6000
0 = 22r^2 - 1420r + 6000
22r^2 - 1420r + 6000 = 0
2(11r-50)(r-60) = 0
11r-50 = 0 or r-60 = 0
11r = 50 or r = 60
r = 50/11 or r = 60
r = 4.54545 or r = 60
If r = 4.54545, then r-10 = 4.54545-10 = -5.45455
It makes no sense to have a negative speed. So we ignore the solution r = -5.45455
The only practical solution is r = 60
Because r = 60, r-10 = 60-10 = 50
So the speed on the way to the destination was 60 mph. The speed coming back was 50 mph.
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