Question 1050412
.
the boat goes downstream at 13 kph
upstream it goes three kph
what is the speed of the current?
It is obvious it is ten kph, but how do you put this into an equation?
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<pre>
When the boat goes downstream, its speed relative to the river's bank is the sum of the boat rate in still water and the current rate.


When the boat goes upstream, its speed relative to the river's bank is the difference of the boat rate in still water and the current rate.


This it the key for solving "Upstream-Downstream" Travel and Distance problems.


So, your equations are

u + v = 13,    (1)    (u is the boat speed in still water,
u - v =  3.    (2)     v is the currebt speed.)

Add the equations (1) and (2). You will get

2u = 13 + 3 = 16.   Hence,  u = {{{16/2}}} = 8 kph.

Then you can easily find v from (1): v = 13-8 = 5.

<U>Answer</U>.  The current speed is 5 kph.  The boat speed in still water is 8 kph.
</pre>

See the lessons 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems.lesson>Wind and Current problems</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/More-problems-on-upstream-and-downstream-round-trips.lesson>More problems on upstream and downstream round trips</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems-solvable-by-quadratic-equations.lesson>Wind and Current problems solvable by quadratic equations</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Selected-problems-from-the-archive-on-a-boat-floating-Upstream-and-Downstream.lesson>Selected problems from the archive on the boat floating Upstream and Downstream</A> 

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