An airplane has airspeed of 724 k/h at bearing of North 30 degrees East. If the wind velocity is 32 from west. find the ground speed and direction of plane.
Teacher gives solution of 740.5 k/hr and 57.9 degrees.
Can anyone show steps?
To find the ground speed and direction of the plane, we need to break down the velocity of the plane and the wind velocity into their horizontal and vertical components.
Given:
Airspeed of the plane = 724 km/h
Bearing of the plane = North 30 degrees East
Wind velocity = 32 km/h from the west (or directly east)
Step 1: Find the horizontal and vertical components of the plane's velocity:
To do this, we can use trigonometry.
Horizontal component = Airspeed * cos(Bearing)
Vertical component = Airspeed * sin(Bearing)
Horizontal component = 724 * cos(30°)
Vertical component = 724 * sin(30°)
Step 2: Find the horizontal and vertical components of the wind velocity:
In this case, since the wind is blowing directly east (from the west), the horizontal component of the wind velocity will be 32 km/h, and the vertical component will be 0 km/h.
Horizontal component of wind velocity = 32 km/h
Vertical component of wind velocity = 0 km/h
Step 3: Add up the horizontal and vertical components of the plane and wind velocity separately:
Horizontal component of total velocity = Horizontal component of plane's velocity + Horizontal component of wind velocity
Vertical component of total velocity = Vertical component of plane's velocity + Vertical component of wind velocity
Step 4: Calculate the magnitude and direction of the total velocity:
Magnitude of the total velocity (ground speed) = √(Horizontal component of total velocity)^2 + (Vertical component of total velocity)^2
Direction of the total velocity = tan^(-1)(Vertical component of total velocity / Horizontal component of total velocity)
Step 5: Substitute the values into the formulas and solve:
Horizontal component of total velocity = 724 * cos(30°) + 32
Vertical component of total velocity = 724 * sin(30°) + 0
Magnitude of the total velocity = √((724 * cos(30°) + 32)^2 + (724 * sin(30°))^2)
Direction of the total velocity = tan^(-1)( (724 * sin(30°)) / (724 * cos(30°) + 32) )
Calculating these values will give you the ground speed of 740.5 km/h and the direction of 57.9 degrees, as provided by the teacher's solution.
convert headings (not bearings!) to x-y coordinates and add them up:
724@30° = (362,627)
32@90° = (32,0)
addthem up and you have
(394,627)
√(394^2+627^2) = 740.5
90-arctan(627/394) = 32.2°
Note that the proposed solution of 57.9° is measured from the x-axis, so it is not a true compass heading.