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Question 1085270: Brian decides to start training for swimming in a river. The current in the river is
4km/hr. If he swims upstream 2 km and then back downstream to where he started in 3
hours, what is his swimming speed?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let
x = Brian's swimming speed in still water (no current)
t1 = time it takes to swim against the current
t2 = time it takes to swim with the current
Make a table to get
| Distance | Rate | Time | | Against Current | 2 | x-4 | t1 | | With Current | 2 | x+4 | t2 |
Against the current, we can say
d = r*t
2 = (x-4)*t1
t1 = 2/(x-4)
with the current, we can say
d = r*t
2 = (x+4)*t2
t2 = 2/(x+4)
The two time values (t1 and t2) add to 3 hours, so
t1 + t2 = 3
2/(x-4) + 2/(x+4) = 3
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Let's solve for x
2/(x-4) + 2/(x+4) = 3
[2/(x-4) + 2/(x+4)](x-4)(x+4) = 3(x-4)(x+4) ... multiply both sides by (x-4)(x+4)
2(x+4)+2(x-4) = 3(x-4)(x+4)
2x+8 + 2x-8 = 3(x^2-16)
4x = 3x^2-48
4x-4x = 3x^2-48-4x
0 = 3x^2-4x-48
3x^2-4x-48 = 0
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Use a calculator or the quadratic formula to solve 3x^2-4x-48 = 0 to get the solution set {x = -3.388508353532, x = 4.721841686865} which are approximate values.
Toss out the negative x value because negative speeds don't make sense.
The only practical solution is approximately 4.721841686865 which is the final answer. The units are in km per hour.
So this means that Brian's speed in still water (with no current) is approximately 4.721841686865 km/hr.
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