Question #3d423

2 Answers
Jan 16, 2018

641.67 m/(mts)

Explanation:

Suppose,the speed of the hawk in still air is v m/(mts)

So,while going against wind flow its net velocity becomes (v-u) m/(mts) where, u is the velocity of wind.

So,if in this process it goes for a distance of x meters, time taken (t) will be, x/(v-u)

Again when it goes in the direction of wind,its net velocity becomes (v+u) m/(mts)
Hence, here time required (t1) to travel the same distance will become x/(v+u)

Now,given that x = 24 , t = 45 , t1 = 32
Solving it we get, v=641.67 m/(mts)

Jan 16, 2018

Let the speed of the hawk in still air be v_hkm/hr and the speed of wind be v_wkm/hr

Hence time taken by the hawk to cover a distance of 24 km flying against the wind will be 24/(v_h-v_w)hr

By the problem

24/(v_h-v_w)=45/60=3/4hr

=>v_h-v_w=32.........(1)

Again in return journey the time taken by the hawk to cover same distance flying with the wind is given by

24/(v_h+v_w)=32/60=8/15hr

=>v_h+v_w=45..........(2)

Adding (1) and (2) we get

2v_h=77

=>v_h=77/2=38.5km/hr