Question

An airplane is flying with velocity of 70km\hr in north east direction .The wind is blowing 30km\hr from north to south.What is the resultant displacement of the aeroplane in 4 h

Answer 1

The resultant displacement of the airplane in 4 hours is 212.8 km.

Step-by-step explanation:

The components of the airplane's velocity and wind's velocity are:

Airplane:

`v_{a_(x)} = v_(a)cos(45) = 70 km/h*cos(45) = 49.50 km/h`

`v_{a_(y)} = v_(a)sin(45) = 70 km/hsin(45) = 49.50 km/h`

Wind:

`v_{w_(x)} = 0`

`v_{w_(y)} = v_(w) = -30 km/h`

Now, to know the new velocity of the airplane we to find the result vector:

`v_(x) = v_{a_(x)} + v_{w_(x)} = 49.50 km/h + 0 = 49.50 km/h`

`v_(y) = v_{a_(y)} + v_{w_(y)} = 49.50 km/h - 30 km/h = 19.50 km/h`

Now, the magnitude of the new speed of the airplane is:

`v_(a) = \sqrt{v_(x)^(2) + v_(y)^(2)} = \sqrt{(49.50 km/h)^(2) + (19.50 km/h)^(2)} = 53.20 km/h`

Finally, after 4 hours the resultant displacement of the airplane is:

`x = v*t = 53.20 km/h*4 h = 212.8 km`

Therefore, the resultant displacement of the airplane in 4 hours is 212.8 km.

I hope it helps you!