Question

An airplane pilot sets a compass course due west and maintains an air speed of 240 km/h. After flying
for
h, he finds himself over a town that is 150 km west and 40 km south of his starting point. The wind
2
velocity (with respect to ground) is :
(A) 100 km/h, 37° W of S
(C) 120 km/h, 37° W of S
(B) 100 km/h, 37° S of W
(D) 120 km/h, 37° S of W​

Answer 1

Answer:

Distance=240km/hr*$\frac{1}{2}$hr

              =120km

$(V_{w}t) ^{2} =(30)^{2} +(40)^{2}$

$(V_{w}t)=\sqrt{2500}$

$V_{w} =100km/hr(S-W)$

Answer 2

Answer:

100 km/hr and 15 degree west of south.

Explanation:

Since the distance covered is 150km west and 40 km south hence applying Pythagoras theorem we will get the displacement as d^2 = 150^2 + 40^2 which on solving the value of d will be 155 km.

Now, to find the angle x we have tanx = 40/150 which will be x= 15 west of south.

So, the velocity of the pilot with respect to the ground is 155/0.5 since he flew for 1/2 hr.

Again by using cosine law we will have vectors as v3=v1^2 + v2^2 - 2v1v2cosx so on putting v1 as 240 and v2 as 310 we will get the value of the velocity with respect to the ground as 100km/hr.