Question

An airplane pilot wants to fly from city A to city B which is 1000 km due north of city A. The speed
of the plane in still air is 500 km/hr. The pilot neglects the effect of the wind and directs his plane
due north and 2 hours later find himself 300km due north-east of city B. The wind velocity is
(A) 150km/hr at 45°N of E
(B) 106km/hr at 45°N of E
(C) 150 km/hr at 45°N of W
(D) 106 km/hr at 45°N of W​

Answer 1

Answer:

$\color{red}{Mark\: as\: Brainlist}$

Answer 2

The wind velocity is 150 km/hr at 45° N of E.

(A) is correct option.

Explanation:

Given that,

Distance = 1000 km due north of city A

Speed of the plane in air  = 500 km/hr

Times = 2 hours

We need to calculate the wind velocity

Using formula of time

$t=t'$

$\dfrac{d}{v}=\dfrac{d'}{v'}$

$\dfrac{d}{d'}=\dfrac{v}{v'}$

Where, d' = distance city from A to B

d = distance of city B

v = wind velocity

v'=velocity of plane

Put the value into the formula

$\dfrac{300}{1000}=\dfrac{v}{500}$

$v=\dfrac{300}{1000}\times500$

$v=150\ km/hr at 45^{\circ}N\ of E$

Hence, The wind velocity is 150 km/hr at 45° N of E.

Learn more :

Topic : velocity

https://brainly.in/question/21982