# Question #3d423

Jan 16, 2018

$641.67 \frac{m}{m t s}$

#### Explanation:

Suppose,the speed of the hawk in still air is $v \frac{m}{m t s}$

So,while going against wind flow its net velocity becomes $\left(v - u\right) \frac{m}{m t s}$ where, $u$ is the velocity of wind.

So,if in this process it goes for a distance of $x$ meters, time taken $\left(t\right)$ will be, $\frac{x}{v - u}$

Again when it goes in the direction of wind,its net velocity becomes $\left(v + u\right) \frac{m}{m t s}$
Hence, here time required $\left(t 1\right)$ to travel the same distance will become $\frac{x}{v + u}$

Now,given that $x = 24 , t = 45 , t 1 = 32$
Solving it we get, $v = 641.67 \frac{m}{m t s}$

Jan 16, 2018

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 $\frac{24}{{v}_{h} - {v}_{w}}$hr

By the problem

$\frac{24}{{v}_{h} - {v}_{w}} = \frac{45}{60} = \frac{3}{4}$hr

$\implies {v}_{h} - {v}_{w} = 32. \ldots \ldots . . \left(1\right)$

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

$\frac{24}{{v}_{h} + {v}_{w}} = \frac{32}{60} = \frac{8}{15}$hr

$\implies {v}_{h} + {v}_{w} = 45. \ldots \ldots \ldots \left(2\right)$

Adding (1) and (2) we get

$2 {v}_{h} = 77$

$\implies {v}_{h} = \frac{77}{2} = 38.5$km/hr