Classic River-boat problem 

 Q.A river 400m wide is flowing at the rate of 2 m/s.A boat is sailing at the rate of 10m/s with repect to the water in a direction perpendicular to the river. a) Find the time taken by the boat to reach the  opposite bank. (b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?

solution :

The component of the resultant velocity which is responsible for causing the boat to cross the river has magnitude 10m/s. 

So time taken to cross the river is 400/10=40s.

Now the horizontal displacement of the boat is caused by the velocity of river

Distance from point directly opposite to the point from where boat started

  = 40 x 2 =80m  

u can use vector method also. For example : 

if u be the velocity of river,& u' be the velocity of boat, 

then the net velocity of boat = u i + u' j = 2 i  + 10 j 

now the time taken for boat to cross the river, 

= distance / velocity in y-direction 

= 400/ 10 = 40 sec 

distance travelled by boat in horizontal direction, will be purely due to the river's velocity, thrfore drift   

= 2 X 40 = 80  m

consider another problem :

A river flows west to east at a speed of 5m per minute. A man on the south bank can swim at the rate of 10m per minute in still water. In which direction should he swim so as to cross the river 1) in the smallest possible time 2) along the shortest path

Solution

Take the normal NSEW directions (N up, E right, etc). Assume XY axes such that+ve X is along east and +Y along north. Let the origin of the coordinate system be at a point on the south shore from where the man starts.

Speed of man wrt water, v_MW = 10 m/min

Velocity of water wrt to earth, v_WE = 5 i m/min

Let velocity of man wrt eath be v_ME

Let the width of the river be d.

(a) Let the man try to swim in a direction making with the X axis.

v_MW = 10(cos i + sin j)

v_ME = v_MW + v_WE  = (10cos + 5) i + 10sin j

Time taken by the man to cross, T = d/10sin

T is minimum if sin is maximum, ie, = /2

Hence the man should swim perpendicular to the current if he is to reach in shortest possible time.



(b) The shortets possible path is pependicular to the river. For the man to follow this path, v_ME must not have any velocity component along X axis.

10cos + 5 = 0

= 120⁰

Hence the man should swim in a direction making 120⁰ the +ve X direction