An aeroplane has to go from a city A to another city B 500 km away in which B is north of A.The wind is blowing towards east at a speed of 20m/s.The air speed of aeroplane is 40 m/s.The direction in which the pilot should head the plane to reach the point B is
To determine the direction in which the pilot should head the plane to reach point B, we need to consider the velocity vectors of the plane and the wind.
First, let's break down the velocities into their respective components. Since the wind is blowing towards the east, we can represent its velocity as (20 m/s, 0 m/s) in terms of its x and y components.
Since the plane's airspeed is 40 m/s, the velocity of the plane relative to the air is (40 m/s, 0 m/s).
To determine the direction the pilot should head, we need to find the resultant velocity of the plane, which is the vector sum of the velocity of the plane relative to the air and the velocity of the wind.
To calculate the resultant velocity, we add the x and y components separately:
V_resultant_x = V_plane_x + V_wind_x
= 40 m/s + 20 m/s
= 60 m/s
V_resultant_y = V_plane_y + V_wind_y
= 0 m/s + 0 m/s
= 0 m/s
The resultant velocity is (60 m/s, 0 m/s).
Since the wind is blowing towards the east, we need to subtract its x-component from the total x-component of the velocity to compensate for the effect of the wind.
Therefore, the pilot should head the plane towards the east, or in the direction opposite to the x-component of the wind.
In conclusion, the pilot should head the plane towards the west to reach point B, as B is north of A and there is a strong wind blowing towards the east.
draw a diagram. This is nice and easy, since you have a right triangle. The distance traveled by the plane in t hours must form the hypotenuse of the triangle, so
(40t)^2 = (20t)^2 + 500^2
t = 25/√3 = 14.434 hours.
Now you can figure the distance and the angle θ (west of north) such that tanθ = 20t/500