an aircraft with an airspeed of 295km/hr is heading south. wind blowing from East at 30km/hr is against the aircraft.
1) sketch the information clearly showing all angles and vectors.
2) calculate the resultant speed and direction of the aircraft
2. X = -30 km/h = velocity of the wind.
Y = -295 km/h.=velocity of the aircraft.
Q3.
tan A = y/x = -295/-30 = 9.83333.
A = 84.2o S. of W. = direction.
Vr = y/sin A=-295/-sin84.2 = 296.5 km/h.
To answer your question and sketch the information clearly, we need to use vector addition. Here's how we can approach it:
1) Sketching the Information:
Start by drawing a compass rose on a piece of paper or a digital drawing tool. Label the cardinal directions – North, South, East, and West. Then, draw an arrow pointing towards the south to represent the direction of the aircraft. Add another arrow pointing towards the east to represent the direction of the wind.
Now, label the airspeed of the aircraft, which is 295 km/hr, near the south arrow. Similarly, label the wind speed, which is 30 km/hr, near the east arrow. To indicate that the wind is against the aircraft's motion, draw crossed lines (or a small "X") across the east arrow.
2) Calculating the Resultant Speed and Direction:
To calculate the resultant speed and direction of the aircraft, we need to find the vector sum of the aircraft's airspeed and the wind speed. Let's break down the vector components:
The airspeed of the aircraft towards the south can be represented as a vector with magnitude 295 km/hr in the south direction.
The wind blowing from the east can be represented as a vector with magnitude 30 km/hr in the east direction.
To find the resultant vector (or the vector sum), we add the two vectors together. Since the original vectors are at right angles to each other, we can use the Pythagorean theorem to find the magnitude of the resultant vector:
Resultant magnitude = √(295^2 + 30^2) km/hr
= √(87025 + 900) km/hr
= √(87925) km/hr
≈ 296.5 km/hr
To find the direction of the resultant vector, we can use trigonometry. The angle between the resultant vector and the south direction can be found by taking the inverse tangent of the wind speed divided by the airspeed:
Resultant angle = tan^(-1)(30/295)
≈ 5.83° east of south
Therefore, the resultant speed of the aircraft is approximately 296.5 km/hr, and it is heading 5.83° east of the south direction.
I hope this explanation helps you with both sketching and calculating the information!