SOLUTION: the cross-country team left school and trotted at 6 miles per hour. then they ran back to the school at 8 miles per hour. if the total trip took 7 hours. how far was the clubhouse?

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: the cross-country team left school and trotted at 6 miles per hour. then they ran back to the school at 8 miles per hour. if the total trip took 7 hours. how far was the clubhouse?      Log On

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Question 9992: the cross-country team left school and trotted at 6 miles per hour. then they ran back to the school at 8 miles per hour. if the total trip took 7 hours. how far was the clubhouse?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You can use the formula: d = rt Distance = rate X time.
Let's do the calculation for the outbound leg first.
d1+=+%28r1%29%28t1%29
d1+=+6+mph%28t1%29
Now the inbound leg:
d2+=+%28r2%29%28t2%29
d2+=+8+mph%28t2%29
Ok, we know that the outbound distance (d1) is the same as the as the inbound distance (d2).
We also know that the total time (t1 + t2) is 7 hours. We can write this as:
t1+=+7+-+t2
Now, setting d1 = d2 and substituting (7-t2) for t1, we can write:
6+mph%287-t2%29+=+8+mph%28t2%29 Simplify and solve for t2
42+-+6%28t2%29+=+8%28t2%29 Add 6(t2) to both sides.
42+=+14%28t2%29 Divide both sides by 14.
t2+=+3+hrs
But, we need the distance, d1 or d2 (they are the same). Let's find d2.
d2+=+%28r2%29%28t2%29
d2+=+8+mph%283+hrs%29
d2+=+24+miles
The distance to the club house is 24 miles