SOLUTION: Jen’s boat cruised 45 miles upstream and 45 miles back in a total of 8 hr. The speed of the river is 3 mph. Find the speed of the boat in still water.
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Question 285655: Jen’s boat cruised 45 miles upstream and 45 miles back in a total of 8 hr. The speed of the river is 3 mph. Find the speed of the boat in still water.
I need help with this word problem.
Thank you.
Cynthia Found 3 solutions by nerdybill, stanbon, richwmiller:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Jen’s boat cruised 45 miles upstream and 45 miles back in a total of 8 hr. The speed of the river is 3 mph. Find the speed of the boat in still water.
.
You will need to apply the "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
However, for this problem we are going to solve for t:
d = rt
d/r = t
t = d/r
.
Let x = speed of boat in still water
then
x+3 = speed going downstream
x-3 = speed going upstream
.
"time going upstream" + "time going downstream" = 8
45/(x+3) + 45/(x-3) = 8
multiplying both sides by (x+3)(x-3):
45(x-3) + 45(x+3) = 8(x+3)(x-3)
45x - 135 + 45x + 135 = 8(x^2-9)
90x = 8x^2-72
0 = 8x^2-90x-72
0 = 4x^2-45x-36
Factoring the right:
0 = (4x+3)(x-12)
x = {-3/4 , 12}
Well we can toss out the negative answer leaving:
x = 12 mph (speed in still water)
You can put this solution on YOUR website! Jen’s boat cruised 45 miles upstream and 45 miles back in a total of 8 hr. The speed of the river is 3 mph. Find the speed of the boat in still water.
-------------------
Upstream DATA:
distance = 45 miles ; rate = b-3 mph ; time = d/r = 45/(b-3) hrs.
------------------------------
Downstream DATA:
distance = 45 miles ; rate = b+3 mph ; time = d/r = 45/(b+3) hrs
------------------------------
Equation:
time + time = 8 hr
45/(b-3) + 45/(b+3) = 8
Multiply thru by (b^2-9) to get:
45(b+3) + 45(b-3) = 8(b^2-9)
90b = 8b^2 - 72
Rearrange:
8b^2 - 90b - 72 = 0
4b^2 -45b - 36 = 0
--------------------------
(b-12)(4b+3) = 0
Positive solution:
b = 12 mph (boat speed in still water)
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Cheers,
Stan H.
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You can put this solution on YOUR website! Just curious.
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