Melody

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Melody  Feb 11, 2022
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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.
Mar 14, 2014
 #68
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Hi all,
I have recovered from my yesterday's collapse. I thank those people who voiced there support (by private message). I really needed to hear it and I am very grateful.
It is thought that the best way to deal with our 'Professional Troll' is to ignore his/her posts. No response = No fun for the troll.
I guess this forum can cope with 1 rude adult without falling in a heap.

Today was a standard busy mid-week day. Fantastic answers were provided by Reinout-g, Alan, Dms, Samgelinas, Ironn, Mathcalc, ChrisP and unknown.
NerdPro also answered a question but it was answer only with no explanation so it was not very helpful. Sorry NerdPro, I am pleased that you answered but you do need to explain the answers that you give.

ChrisP, it is great that you are answering so many questions. We'd really like you to consider joining up. There are a number of benefits and we would love you to join us formally.
Look here to find out what benefits there could be for you.
http://web2.0calc.com/questions/become-a-member-why-would-you-want-to

Alan sent an email to Andre Massow yesterday. Andre responded and said that moderator rights will be a feature in the new forum design. Andre also said that he was hoping to publish a new Beta version upgrade this weekend so that we can all evaluate it. I am not to sure what that means. I guess we will find out.

Oh there was a question today with notation that I haven't seen before. It was in here http://web2.0calc.com/questions/quads It was just m<ABC which meant measurement of angle ABC. I had no idea what the 'm' meant. Chris knew what it meant. Is it normal notation? Could any comments please be posted on the end of the relevant thread. Please don't post here.

That's it for this Wed/Thurs,
Thank you everyone,
Melody.
Mar 13, 2014