-------------> V river = 5m/min     
V man = 10 m/minsuppose width of river is x mts
Case 1
if he travels due north
then his resultant velocty ( due to velocity of downstream of water) acts 30 north of east and the distane to be covered to reach the other bank increses to = x/cos 30
i.e., 2x /31/2 and resultant velocity is 11.2 m/min
so time taken is x / 9.7 mins
Case 2
if he travels due 30 north of east
then his resultant velocty ( due to velocity of downstream of water) actsdue north and the distane to be covered to reach the other bank will be x mts and resultant velocity is 8.7 m/min
so time taken is x / 8.7 mins
So case 1 takes least time
he should travel due north(a)
 
 
						

							
he should swim due north!!!
his velocity in moving water is equal to his own velocity plus the velocity of the river.
now to cross the river in smallest possible time he should maximise the velocity in the north direction.
suppose he move at an angle of @ from the bank then the component due north is 10sin@m/min. because river velocity is zero in the north direction.
now this velocity due noth will be maximum when @ tends to 90 degrees.
hence he should move due north that is 90 degrees to the bank.
						

							Thus ,the resultant of the Velocity of man in still water (Vm)(=OA)and the Velocity of water (Vw)(=OB) is along OP, and is given by (V)(=OC)In right triangle OCA , we haveSin x= AC/OA=Vw/Vm        =5/10  = 1/2x=30°Therefore, the man should start swimming at an angle of 30° (west of north)
						

							
Yes we should move due north to cover the distance in the shortest time as then the velocity of river and man will get added up and distance would be covered in the shortest time
						

							
If the man crossed the river with a angle @ to examined the river velocity then he should travel due north with velocity vcos@.so time taken to cross the river is d/vcos@.after differentiate both the sides of eqn we get @=0. So he should travel in north direction.