# A man can swim at a speed of 3km/hr in still water;he wants to cross a 500m wide river flowing at 2km/hr.He keeps himself always at an angle of 120 with the river flow while swimming.Find the time he will take to cross the river.At what point on the opposite bank will he arrive?

## Answers

given,

speed of man, Vm = 3km/hr

speed of river, Vr = 2km/hr

angle which the man makes with the direction of motion of river = 120

The resultant velocity of man can be calculated by the parallelogram law of vectors as,

angle between Vr and VR

now time required to cross the river = OB/VR

in trianle ABO, angleAOB = 90o - 79o = 11o

so, OB = AO/cos11o = 0.5/0.981 = 0.509

so **time required = 0.509/2.6 = 0.196hr = 0.196x60x60 = 705.6s**

Now the point where he will reach is B**,** and **the distance of B from A = AB = AOtan11 = 0.097km **

** = 0.097x1000= 97m**

**Answer:**

Time=1/3 hr or 20 min

point on opp. bank/drift=1/6 km

**Explanation:**

Resolving his speed.

At x axis-: -1.5√3 kmph

at y axis: 1.5 kmph

Now,

Width of river=500 m=500/1000 km

speed in y axis=1.5

so, time=width /speed =1/3 hr

net speed in y axis=2-1.5√3=0.5 kmph (approx.)

so drift=net speed in x axis*time=0.5*1/3=1/6km(ans)

hope this helps