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# Pre Calc

-1
121
1

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?

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)   Feb 10, 2020
edited by CPhill  Feb 10, 2020