+318 A motorboat is capable of traveling at a speed of
15
miles per hour in still water. On a particular day, it took
15
minutes longer to travel a distance of
5
miles upstream than it took to travel the same distance downstream. What was the rate of current in the stream on that day?
+129852 Let the rate of the current = R
Note that 15 min = 1/4 hr
So we have that
The time with the current + 1/4 hr = time against the current .....and with the current the rate is 15 + R .....and against the current the rate is 15 - R ......and distance / rate = time
So
5 1 5
______ + ____ = ______ rearrange as
15 + R 4 15 - R
5 5 1
______ - ______ = - ____
15 + R 15 - R 4
5 ( 15 - R) - 5(15 + R) 1
_________________ = - ______
(15 + R) ( 15 - R) 4
-10R -1
________ = ______ cross-multiply
225 - R^2 4
-40R = 225 - R^2
R^2 - 40R - 225 = 0 factor
(R - 5) ( R + 45) = 0
Only the first factor set to 0 gives us the speed of the current
R - 5 = 0
R = 5 (mph)
