a pilot must fly a destination 520 km away in a direction of N30 degree E. The air speed of the plane is 320Km and the wind speed is 85 km/h E20 degree Suth. Find the heading of the plane and the time for the trop.
This question assumes knowledge of vectors and how to solve for a triangle using the sine rule. If these subjects are vague, a revision is in order.
Let the plane start at O (origin) in the x-y plane, x=East, and y=North. Use a piece of graphics paper to help you draw the sketch according to the following directions. Draw it roughly to scale, but it does not have to be accurate, as we will solve the unknowns by trigonometry.
We will consider a time period of 1 hour, hence the plane travels 320 km and the wind, 85 km. We take the vectorial sum of the effect of wind and the plane, the direction of wihci is yet unknown.
Draw a line from O in the direction of the destination, namely N30Em, length about 400 units (km).
Draw the wind vector OA, in the direction E20S and a magnitude of 85 km/h.
From A, draw a vector AB, of magnitude 320 and intersecting the line in the N30E direction at B.
If we isolate the triangle OAB, we have:
Angle BOA = (90-30)+20 = 80°
OA = 85 km
Using sine rule,
sin(B)/85 = sin(80)/320
solve to get B=15.1644°
Since angles O + A + B =180,
we solve for A=180-O-B = 84.8356°.
Using sine rule again,
OB/sin(A) = 320/sin(80)
we solve for OB=324 km/hr approx.
Heading: NθE where θ=30-∠B=15° approx.
From A, draw a vector AB, of magnitude 320 and intersecting the line between O and destination.
If we isolate the triangle OAB, we have:
Angle BOA = (90-30)+20 = 80°
OA = 85 km
OB = 320 km
To find the heading of the plane, we need to consider the effect of the wind on the plane's direction.
1. Find the ground speed of the plane:
The ground speed is the resulting speed of the aircraft when its airspeed is combined with the wind speed. We can calculate the ground speed using vector addition.
The airspeed of the plane is 320 km/h, and the wind speed is 85 km/h E20°S. We can break down the wind speed into its north-south (N-S) and east-west (E-W) components.
Wind speed N-S component = Wind speed * sin(wind direction)
= 85 km/h * sin(20°S)
≈ -29.00 km/h (negative sign indicates south direction)
Wind speed E-W component = Wind speed * cos(wind direction)
= 85 km/h * cos(20°S)
≈ 80.59 km/h
Now, we can add the airspeed and wind speed components to find the ground speed.
Ground speed N-S component = Airspeed N-S component + Wind speed N-S component
= 320 km/h * sin(30°E) + (-29.00 km/h)
≈ 166.28 km/h (positive sign indicates north direction)
Ground speed E-W component = Airspeed E-W component + Wind speed E-W component
= 320 km/h * cos(30°E) + 80.59 km/h
≈ 359.26 km/h
Using these ground speed components, we can find the magnitude and direction of the ground speed.
Magnitude of the ground speed (|V|) = √((Ground speed N-S component)^2 + (Ground speed E-W component)^2)
≈ √((166.28 km/h)^2 + (359.26 km/h)^2)
≈ 397.15 km/h
Direction of the ground speed = arctan((Ground speed N-S component) / (Ground speed E-W component))
≈ arctan(166.28 km/h / 359.26 km/h)
≈ 23.35°
Therefore, the heading of the plane is N23.35°E.
2. Find the time for the trip:
To find the time for the trip, we can use the formula:
Time = Distance / Ground Speed
Distance = 520 km
Ground Speed = 397.15 km/h (as calculated above)
Time = 520 km / 397.15 km/h
≈ 1.31 hours
Therefore, the time for the trip is approximately 1.31 hours.