a twin engine plane flies due south with an air speed of 335km/h. There is a wind blowing at 75km/h to the east relatve to the ground. what are the plane's speed and direction relative to the ground
speed=sqrt(335^2+75^2)
direction: arcTan75/335 E of S
To find the plane's speed and direction relative to the ground, we can use vector addition.
1. First, let's represent the southward airspeed of the plane as a vector. Since the plane is flying due south, the vector representing the airspeed will have a magnitude of 335 km/h in the south direction (-y direction).
2. Next, let's represent the wind speed as a vector. The wind is blowing to the east, so the vector representing the wind will have a magnitude of 75 km/h in the east direction (+x direction).
3. Now, we can add these two vectors to find the resultant vector, representing the plane's velocity relative to the ground.
a. To add the vectors, we need to break them down into their x and y components. The airspeed vector has a magnitude of 335 km/h in the -y direction, so its y component is -335 km/h. The wind vector has a magnitude of 75 km/h in the +x direction, so its x component is +75 km/h.
b. Now, we can add the x components and the y components separately. Adding the x components (+75 km/h) results in a total of +75 km/h in the east direction. Adding the y components (-335 km/h) results in a total of -335 km/h in the south direction.
c. The resultant vector has an x component of +75 km/h and a y component of -335 km/h.
4. Finally, we can find the magnitude and direction of the resultant vector using the Pythagorean theorem and trigonometry.
a. The magnitude of the resultant vector is given by the square root of the sum of the squares of the x and y components. Magnitude = sqrt((75 km/h)^2 + (-335 km/h)^2) = sqrt(5625 km^2/h^2 + 112225 km^2/h^2) = sqrt(117850 km^2/h^2)
b. The direction of the resultant vector can be found using the inverse tangent function. Direction = arctan(y component / x component) = arctan((-335 km/h) / (75 km/h))
c. Use a calculator to find the inverse tangent of -335/75, which gives us approximately -77.6 degrees.
Therefore, the plane's speed relative to the ground is approximately sqrt(117850) km/h, and its direction is approximately 77.6 degrees south of east.