Question

A bkat which has a speed of 5km/hr in stoll water 34.68 minutea

Answer 1

      A               B

               ^                ^

                *              1

                   *           1

                 u    *       1    r

                          *    1

                              *1------------------------à  C  

                                 D               v

 sorry   i dont have a paint brush.......so manage with this fig.

 

Let  angle ADB = x

         Angle BDC =90

Vector AD = velocity of boat in still water  ( u)

Vector  BD =  resultant velocity ( shortest path)   (  r)

* for shortest path , resultsnt velocity shud b perpendicular to the bank

Vector  DC = velocity of river (v)

 

     r= 1km/ 15 min = 4 km/hr -------------(1)

    Cos x =  BD/AD

AD = 5 Km /hr *  15 /60 Hr  = 5/4 km

Cos x = 1/ (5/4) = 4/5

 So Sin x = 3/5  ---------------  (2)

Now using r2 = u2+ v2 + 2uvcosx

                 16  = 25 +  v2   + 10vcos ( 90+ x)

                  16  = 25 +  v2  -  10 v sinx

     putting value of sin x            

ð      v2 ?6v +9 =0

ð      solving the quadratic eqn , we get

ð      v = 3 km/hr

may it's help you