fiz khan: :geek: A man swims at a speed of 50m/min in still water,swims 100m against the current and 100m with the current. If the difference between the two times is 3 min 45 s find the speed of the current
NOTE: IT WILL BE MUCH EASIER TO FOLLOW WHAT I AM SAYING IF YOU RIGHT ALL YOUR FRACTIONS UPRIGHT.
Let the current be y metres/minute
When he swim against the current his speed is (50-y) m/min. Which is equaivalent to saying that in 1 minute he will swim (50-y) metres.
So the rate can be expressed as 1/(50-y) minutes/metre OR 1minute / [(50-y)metres]
Now
1minute / [(50-y)metres] x 100metres = 100/(50-y) minutes {The metres on the bottom of the fraction have cancelled with the metres on the top of the fraction}
(*Teachers don't teach it this way to very often but I don't know why not, it is the easiest way way to deal with any question involving rates)
When he swims with the current his speed is (50+y)m/minute. Which can be expressed as 1minte/[(50+y)metres]
1minte/[(50+y)metres] x 100m = 100/(50+y) minutes
You are told that
100/(50-y) - 100/(50+y) = 3.75
the easiest way to handle this equation is to multiply both sides by (50-y)(50+y)
(50-y)(50+y) [ 100/(50-y) - 100/(50+y) ] = (50-y)(50+y)* 3.75
100(50+y) - 100(50-y) =3.75 (2500-y
2)
5000+100y-5000+100y = 9375 - 3.75y
2 200y = 9375 - 3.75y
2 3.75y
2+200y-9375=0 [You can solve this using the quadratic formula but I will use the site calculator]
[input]3.75y^2+200y-9375=0[/input]
The negative answer can be discounted so the speed of the current must be 30metres/minute
You should check this answer by fining out how long it takes to swim 100m at 80m/min and 100m at 20m/min and finding the difference.
+118723 
