THE CLASSIC RIVER - BOAT/SWIMMER PROBLEM

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THE CLASSIC RIVER - BOAT/SWIMMER PROBLEM

MANY A TIMES THE APPLICATION OF VECTOR IS QUITE DIFFICULT FOR THOSE WHO HAVE LITTLE OR NO PRIOR KNOWLEDGE OF IT.SO HERE I HAVE TRIED TO BRING OUT A GENERAL EQUATION FOR THE DRIFT AND ANGLE THAT CAN BE USED BY THE READERS TO SOLVE

?CROSSING THE RIVER? TYPE OF QUESTION.

CONSIDER A GENERAL CASE WHEN THE SPEED OF A BOAT OR SWIMMER IS ?v? WITH RESPECT TO STILL WATER. ALSO THE SPEED OF THE STREAM IS ?u?. THE RIVER IS ?l? UNITS WIDE.SUPPOSE THE DRIFT FROM HIS ACTUAL PATH IS x.

THE VELOCITY OF OBJECT IN THAT DIRECTION IS =

STREAM SPEED + COMPONENT OF OBJECT SPEED IN THAT DIRECTION.

w = u + v*cos? where ? is the angle between the swimmer and the Stream velocity

HENCE THE DRIFT IN THAT DIRECTION IS

x = ( u + v*cos? )*t where t is the time taken to cross the river ??.(1)

NOTE : HERE IT IS IMPORTANT TO NOTE THAT IF THE OBJECT IS MOVING AGAINST THE STREAM ?>90. HENCE cos ? IS NEGETIVE . THIS REDUCES THE DRIFT AS IS EXPECTED IN REAL LIFE SITUATION.

l = v*t*sin ?

t = l/v*sin ? , NOW SUBSTITUTING THIS IN (1) WE HAVE

x = { (u/v*sin ?) + cot ? }

THESE VALUES ARE GENERALLY GIVEN IN THE QUESTION AND HENCE FINDING DRIFT NOW TAKES NO TIME.NOW IF NOW THE VALUE OF u/v OR IF IT CAN BE FOUND NUMERICALLY THEN LET u/v=n .

SO, x = { n*cosec ? + cot ? }*l

HENCE FOR MINIMUM DRIFT ( dx/d? ) = 0;

-n*cosec? ? cosec?*cosec? = 0

OR cosec? = 0 OR cos? = -1/n

BUT cosec? CANNOT BE ZEROAS IT IS NOT IN RANGE (-?,1]u[1, ?)

SO ? = cos?( -1/n ) .................(2)

HENCE THE ANGLE CAN BE FOUND OUT BY JUST KNOWING n.

CAUTION: THE SECOND EQUATION CAN BE USED ONLY WHEN u>v OR ELSE THE VALUE OF cos? WOULD BE OUT OF RANGE.

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THE CLASSIC RIVER - BOAT/SWIMMER PROBLEM