Question 934432: A BOAT CAPABLE OF MOVING IN STILL WATER WITH A SPEED OF 10 KM/HR CROSSES A RIVER 2 KM BROAD FLOWING WITH A SPEED OF 6 KM/HR. FIND THE TIME OF CROSSING BY THE SHORTEST ROUTE, AND FIND THE MINIMUM TIME OF CROSSING
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! first, the shortest route is directly across the 2 km width ...
to cross this way the boat must aim into the current at an angle to negate the speed of the current:
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set up a right triangle with:
h = hypotenuse = boat speed in still water = 10 kph
a = short leg = current speed = 6 kph
x = long leg = effective speed of boat in the current
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h = sqrt( aa + xx )
10 = sqrt( 6*6 + xx )
( 10 = sqrt( 6*6 + xx ) )^2
10*10 = 6*6 + xx
xx = 10*10 - 6*6
x = sqrt( 10*10 - 6*6 )
x = 8 kph
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s = d/t
t = d/s
t = 2/8
t = 0.25 hours = 15 minutes
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answer A:
time to cross by the shortest distance route = 15 minutes
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now, the minimum crossing time ...
the fastest crossing is when the boat aims directly perpendicular to the current:
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set up a right triangle with:
h = hypotenuse = effective speed of boat in the current
a = short leg = current speed = 6 kph
x = long leg = boat speed in still water = 10 kph
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to calculate this crossing time we only need to know x, the speed of the boat perpendicular to the current, which we know is 10 kph:
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t = d/s
t = 2/10
t = 0.20 hours = 12 minutes
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answer B:
the minimum crossing time = 12 minutes
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