Question

a swimmer jumps in a river and swim upto a distance of 80 km and returns back t o its starting position in 8 hours.find average speed and average velocity.​

Answer 1

Answer:

Explanation:

Here θ+α = 90°

θ = 90°-α

 

In ΔOBC, sin θ=sin (90°-α) = cos α

Angle CBO =  α, Hence we can write that:

cos α =BC/OB

cos α =5/10

α =cos-1 (1/2)

α = 60° upstream.

So the swimmer should swim 60º upstream.

 

Let v be the resultant velocity of the swimmer i.e. along OC

begin mathsize 14px style straight v space equals square root of straight v subscript straight s squared minus straight v subscript straight r squared end root straight v equals square root of 10 squared minus 5 squared end root straight v equals 8.6 space km divided by straight h end style

 

ii)

 

Time taken by the swimmer to cross the river of width 100 m

t = 100 m / v

t = 0.1 km / 8.6 km/h

t =0.0116 hr

t =0.696 minutes