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Blazing goIITian

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29 Mar 2008 21:32:27 IST

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simple relative velocity sum

None

A SWIMMER WISHES TO CROSS A 500M WIDE RIVER FLOWING AT 5KM.

HIS SPEED WITH RESPECT TO WATER IS 3KM/HR. THE MAN HAS TO REACH THE OTHER SHORE AT A POINT DIRECTLY ABOVE HIM . IF HE REACHS SOMEWHERE ELSE HE HAS TO WALK TO THE POINT. FIND THE MINIMUM DISTANCE HE HAS TO WALK

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Comments (15)

sharmadon

Cool goIITian

Joined: 14 Mar 2008

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29 Mar 2008 21:57:45 IST

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is it 2500/3? pls confirm..then i shall post my soln..

VARUN RAJ

Blazing goIITian

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29 Mar 2008 21:59:48 IST

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EXELLENT DUDE COULD U PLS EXPLN UR WORKING

GAURAV SHARMA

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29 Mar 2008 22:03:12 IST

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look time to reach the opposite end is d/v sin(theta)..now minimum time is wen sin(theta) is maximum..so min. tim = d/v

the drift is u*t=u*d/v = 2500/3

look out for figure in the next post..

VARUN RAJ

Blazing goIITian

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29 Mar 2008 22:04:32 IST

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CANT UNDERSTAND

GAURAV SHARMA

Hot goIITian

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29 Mar 2008 22:06:29 IST

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Re:simple relative velocity sum

GAURAV SHARMA

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29 Mar 2008 22:07:05 IST

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i hope u have understood now..

VARUN RAJ

Blazing goIITian

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29 Mar 2008 22:07:47 IST

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THE QUSTION IS NOT ABT TIME BUT THE DISTANCE FROM THE POINT ABOVE HIM SHOULD BE MINIMUM

VARUN RAJ

Blazing goIITian

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29 Mar 2008 22:12:53 IST

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SO IM NOT AGNST UR METHOD BUT PLS EXPN UR REASONING

GAURAV SHARMA

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29 Mar 2008 22:15:09 IST

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alright then u need to differentiate..and solve..even then u get theta as pie/2..

the equation is

d ( ud/v cosec(theta) ) / d(theta)

u get theta = pie/2..

when u put the values u get the same condition..

VARUN RAJ

Blazing goIITian

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29 Mar 2008 22:18:19 IST

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SEE I DNT KNOW DIFFRENTIATION
COULD U SOLVE IT BY RELATIVE VELOCITY METHOD
I WILL BE GRATEFUL TO U
CHEERS

GAURAV SHARMA

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29 Mar 2008 22:29:48 IST

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no but for minimum and maximum..we need calculus...

the concept is for minimum or maximum put derivative = 0

then to check for maxima or minima..we see the sign of second derivative..

so perhaps u need to wait..

VARUN RAJ

Blazing goIITian

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30 Mar 2008 10:25:51 IST

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if anyone can solve this sum withouth differentiation pls solve it

VARUN RAJ

Blazing goIITian

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30 Mar 2008 11:07:39 IST

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PLS DUDES SOLVE THIS

BALGANESH

Blazing goIITian

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30 Mar 2008 11:08:10 IST

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drifted distance, x = d/vm*vr
vm = 3*5/18 = 5/6m/s
vr = 5*5/18 = 25/18
x = [500/[5/6] ] * 25/18
= 600 * 25/18
= 2500/3

VARUN RAJ

Blazing goIITian

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30 Mar 2008 11:10:36 IST

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PLS EXPLN

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