+0  
 
-1
121
1
avatar+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?

 Feb 10, 2020
 #1
avatar+111389 
+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)

 

 

cool cool cool

 Feb 10, 2020
edited by CPhill  Feb 10, 2020

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