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

Let us draw a diagram for the given situation. Speed of the boat in still water = 5 km /hr

Width of the river = 1 km

Time taken by the boat to cross the river = 15 minutes = Let the velocity of the river be x km/hr.

Distance covered by the boat in 1.25 km

Because of the flow of the river the boat will move in the direction of AC, which is the shortest possible path for the boat.

∴ AC = 1.25 km

AB (Width of the river) = 1 km

BC is the other bank of the river so the width AB of the river will be perpendicular to BC .i.e., ∠B = 90°

In ΔABC by Pythagoras theorem,

(AB)2 + (BC)2 = (AC)2

⇒ (1)2 + (BC)2 = (1. 25)2

⇒ (BC)2 = (1.25)2 – (1)2

⇒ (BC)2 = 1.5625 – 1

⇒ (BC)2 = 0.5625

⇒ BC = 0.75 km

Which means that distance covered by the river water in 15 minutes is 0.25 km.

∴ Velocity of the river =  • 174

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
• 45
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