Plane A is flying at 600 mph in the northeast direction relative to the Earth. Plane B is flying at 500 mph in the north direction relative to the Earth. What is the direction of motion of Plane B as observed from Plane A?Select one:a.10.1° N of Wb.48.2° N of Wc.52.5° N of Ed.37.5° N of W

To solve this problem, we need to use vector subtraction because we are finding the relative velocity of Plane B from Plane A.

Step 1: Draw the vectors.

• Plane A is flying northeast, which can be broken down into two components: 600 mph east and 600 mph north.
• Plane B is flying north at 500 mph.

Step 2: Subtract the vectors.

• The east component of Plane A's velocity doesn't have a counterpart in Plane B's velocity, so it remains as is: 600 mph east.
• The north component of Plane A's velocity is greater than Plane B's velocity, so when you subtract you get: 600 mph north - 500 mph north = 100 mph north.

Step 3: Find the angle.

• We now have a right triangle with sides of 600 (east) and 100 (north). We can find the angle using the tangent function: tan(theta) = opposite/adjacent = 100/600 = 0.1667.
• Use the inverse tangent function to find the angle: theta = arctan(0.1667) = 9.46 degrees.

So, the direction of motion of Plane B as observed from Plane A is approximately 9.46 degrees north of east. However, this option is not available in the choices given.

The problem might be in the interpretation of the directions. If we consider the northeast direction as a 45 degrees angle with respect to the north, then we should subtract this angle from the angle we calculated.

So, 45 degrees - 9.46 degrees = 35.54 degrees.

Again, this option is not available in the choices given.

Therefore, there seems to be a mistake either in the problem or in the provided choices.

To solve this problem, we need to use vector subtraction because we are finding the relative velocity of Plane B from Plane A.

Step 1: Draw the vectors.

• Plane A is flying northeast at 600 mph. This can be broken down into two components: 424.3 mph east and 424.3 mph north (since the angle is 45 degrees, the north and east components are equal).
• Plane B is flying north at 500 mph. This has no east component and a north component of 500 mph.

Step 2: Subtract the vectors.

• The east component of Plane B as observed from Plane A is 0 - 424.3 = -424.3 mph (west).
• The north component of Plane B as observed from Plane A is 500 - 424.3 = 75.7 mph (north).

Step 3: Find the magnitude and direction of the resultant vector.

• The magnitude can be found using the Pythagorean theorem: sqrt((-424.3)^2 + 75.7^2) = 429.3 mph.
• The direction can be found using the arctan function: arctan(75.7 / 424.3) = 10.1 degrees.

Therefore, the direction of motion of Plane B as observed from Plane A is 10.1° N of W. So, the correct answer is a.10.1° N of W.