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?

mharrigan920 Feb 10, 2020

#1**+1 **

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)

CPhill Feb 10, 2020