Question 158338

Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r

Let r=rate (speed) of the man in still water

Time to travel upstream =2/(r-2) (upstream we need to subtract the speed of the current)
Time to travel downstream=2/(r+2) (downstream we add the speed of the current)

Now we are told that the above two times add up to 1 hour, so:

2/(r-2) + 2/r+2)=1  multiply each term by (r-2)(r+2)
2(r+2)+2(r-2)=(r-2)(r+2)  get rid of parens

2r+4 +2r-4=r^2-4 or
4r=r^2-4  subtract 4r from each side

r^2-4r-4=0 quadratic in standard form; solve using the quadratic formula

{{{r = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{r = (4+- sqrt((-4)^2-4*1*(-4) ))/(2) }}} 

{{{r = (4+- sqrt(16+16 ))/(2) }}} 
{{{r = (4+- sqrt(32 ))/(2) }}}
{{{r = (4+- 5.66)/(2) }}}  
We will discount the negative value for r; speed in this problem is positive
{{{r=(4+5.66)/2=9.66/2=4.83}}} kph---speed of man in still water
2/(4.83-2) +2/(4.83+2)=1
2/2.83 +2/6.83=1
0.707+0.293=1
1=1

Hope this helps---ptaylor