Question 233476
Travis paddles his kayak in the harbor at Morro Bay, California, where the incoming tide has caused a current in the water.
 From the point where he enters the water, he paddles 2 miles against the current, then turns around and paddles 2 miles back to where he started. 
His average speed when paddling with the current is 3 miles per hour faster than his speed against the current
 If the complete trip (out and back) takes him 1.4 hours, find his average speed when he paddles against the current.
:
s = av speed against the current
then it says, av speed with the current is 3 mph faster than against, therefore
(s+3) = av speed with the current
:
Write a time equation: time = {{{dist/speed}}}
Time with + time against = 1.4 hrs
{{{2/((s+3))}}} + {{{2/s}}} = 1.4
Multiply equation by s(s+3), results:
2s + 2(s+3) = 1.4s(s+3)
:
2s + 2s + 6 = 1.4s^2 + 4.2s
:
4s + 6 = 1.4s^2 + 4.2s
;
Arrange as a quadratic equation
1.4s^2 + 4.2s - 4s - 6 = 0
:
1.4s^2 + .2s - 6 = 0
:
Use the quadratic formula to find the solution for s
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation x=s; a=1.4; b=.2; c=-6
{{{s = (-.2 +- sqrt(.2^2 - 4*1.4*-6 ))/(2*1.4) }}}
:
{{{s = (-.2 +- sqrt(.04 - (-33.6) ))/(2.8) }}}
:
{{{s = (-.2 +- sqrt(.04 + 33.6 ))/(2.8) }}}
:
{{{s = (-.2 +- sqrt(33.64))/(2.8) }}}
the positive solution
{{{s = (-.2 + 5.8)/(2.8) }}}
s = {{{5.6/2.8}}}
s = 2 mph against the current
:
:
Check solution by finding the times. (speed with the current: 2+3 = 5 mph}
{{{2/5)}}} + {{{2/2}}} = 1.4