THE FLOW SPEED OF WATER VARIES WITH THE DISTANCE FROM ITS STRAIGHT BANK AS V=(1m^1/2s).z^{1/ 2} where z is the perpendicular distance in m from the bank A.SWIMMER ENTERS THE RIVER AT A POINT A ON THE BANK AND SWIMS AT 2m/s, IN A DIRN NORMAL TO RIVER FLOW. WHAT IS HIS DRIFT DOWNSTREAM BY THE TIME HIS NORMAL DISTANCE FROM THE bank IS 20m?

Dear vardaan

V=z^{1/ 2}

velocity of man =2 m/sec

time taken by man to swim 20 m in normal direction t= 20/2 =10 sec

and time taken be man to swim z distance in normal direction t= z/2............1

now consider motion from z to z+dz distance time taken by man to travel z to z+dz postion is dt

and for this small distance dz ,velocity of river assumed to be constant V =z^{1/ 2}

so distance moved in downstreem direction in time dt is

ds =z^{1/ 2} dt

ds =(2t)^1/2 dt  from equation 1

∫ds =∫(2t)^1/2 dt   

S =√2 t^3/2 /3/2  limit o to 10 

   =2√2/3 t^3/2    limit o to 10

   =40√5/3

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

 We are all IITians and here to help you in your IIT JEE preparation.

All the best.
 
Regards,
Askiitians Experts
Badiuddin