Question

a boat which has a speed of 5km/hrs. in still water crosses a river width 1km along the shortest possible path in 15minutes. what is the velocity of river water ?

Answer 1

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

Answer 2

shortest path is 1km

time is 15mins = 1/4 hrs

shortest path means component of velocity perpendicular to river water flow...

So speed = distance/time = 1/(1/4) = 4 km/hr

velocity of boat is 5 km/hr

So component of velocity perpendicular to river flow according to Pythagoras theorem

= underroot (5^2 - 4^2) = 3 km/hr. (answer)

hope it helps....
pls correct me if i m wrong...
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