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_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