Question

Question #3d423

Answer 1

`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)`

Answer 2

Let the speed of the hawk in still air be `v_h`km/hr and the speed of wind be `v_w`km/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=3/4`hr

`=>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=8/15`hr

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

Adding (1) and (2) we get

`2v_h=77`

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