The pilot of an aircraft wishes to fly due

west in a 50.0 km/h wind blowing toward
the south. The speed of the aircraft in the absence of a wind is 228 km/h.

a) How many degrees from west should the
aircraft head? Let clockwise be positive.
Answer in units of degrees.

b) What should the plane’s speed be relative to the ground? Answer in units of km/h

a) The pilot wants to fly due west, so he needs to head in the opposite direction of the wind. Since the wind is coming from the south, he needs to head towards the north. This creates a right-angled triangle with the wind speed as the adjacent side and the speed of the aircraft as the hypotenuse. Using trigonometry, we can find the angle:

tan(angle) = wind speed / aircraft speed

tan(angle) = 50.0 km/h / 228 km/h

angle = tan^(-1)(50.0/228)

angle ≈ 12.5 degrees

b) To find the plane's speed relative to the ground, we can use the Pythagorean theorem. The aircraft's speed acts as the hypotenuse, and the wind speed acts as one of the legs. The other leg will be the plane's speed relative to the ground.

(plane's speed relative to the ground)^2 = (aircraft speed)^2 - (wind speed)^2

(plane's speed relative to the ground)^2 = (228 km/h)^2 - (50.0 km/h)^2

(plane's speed relative to the ground)^2 = 51984 km^2/h^2 - 2500 km^2/h^2

(plane's speed relative to the ground)^2 = 49484 km^2/h^2

(plane's speed relative to the ground) ≈ √49484 km/h

(plane's speed relative to the ground) ≈ 222.4 km/h

Therefore, the plane's speed relative to the ground should be approximately 222.4 km/h.

To determine the heading and speed of the aircraft, we'll need to use vector addition. Here's how you can calculate it step-by-step:

a) Finding the heading angle:
1) Start by drawing a diagram with the east direction labeled as the positive x-axis, and the north direction labeled as the positive y-axis.
2) The wind is blowing toward the south, so its direction is in the negative y-axis (-y direction).
3) With respect to the ground, the speed of the wind is 50.0 km/h in the -y direction.
4) The speed of the aircraft in the absence of wind is 228 km/h.
5) Due to the wind, the actual ground speed of the aircraft will be the vector sum of the aircraft's speed and the wind's speed.
6) Since the aircraft wishes to fly due west (along the -x direction), we'll need to find the angle between -x axis and the resultant vector.

To calculate the angle:
7) Use the trigonometric inverse tangent function (arctan) to find the angle. Specifically, use the formula:
angle = arctan(resultant_y / resultant_x)

To find the resultant x and y components:
8) Since the aircraft is traveling west, its x-component is -228 km/h. The y-component is -50.0 km/h (due to the wind).
9) Use Pythagorean theorem to find the magnitude of the resultant vector (ground speed of the aircraft):
resultant_magnitude = sqrt(resultant_x^2 + resultant_y^2)

b) To determine the plane's speed relative to the ground (resultant_magnitude), you'll need to solve for it using the magnitude found in step 9.

Now let's do the calculations:

a)
resultant_x = -228 km/h
resultant_y = -50.0 km/h

Angle = arctan(-50.0 km/h / -228 km/h)
= arctan(0.2193)
= 12.6 degrees (rounded to one decimal place)

Therefore, the aircraft should head 12.6 degrees clockwise from west.

b)
resultant_magnitude = sqrt((-228 km/h)^2 + (-50.0 km/h)^2)
= sqrt(51984 + 2500)
= sqrt(54484)
= 233.5 km/h (rounded to one decimal place)

Therefore, the plane's speed relative to the ground is 233.5 km/h.

To answer these questions, we can break down the problem into components. The wind is blowing towards the south, and the pilot wants to fly due west.

a) How many degrees from west should the aircraft head?

To determine the angle, we can use trigonometry. We have a right angled triangle, where the vertical side represents the wind's southward velocity, and the horizontal side represents the aircraft's westward velocity. The angle we are interested in is the angle between the direction the aircraft is heading and the westward direction.

We can use the tangent function to find this angle. The tangent of an angle is equal to the ratio of the opposite side to the adjacent side. In this case, the opposite side is the southward wind velocity, and the adjacent side is the westward velocity of the wind. Let's call the angle we're looking for theta.

tan(theta) = opposite/adjacent
tan(theta) = 50 km/h / 228 km/h

Now, let's solve for theta:
theta = arctan(50 km/h / 228 km/h)
theta ≈ 12.68 degrees

So, the aircraft should head approximately 12.68 degrees south of due west.

b) What should the plane's speed be relative to the ground?

To determine the plane's speed, we need to find the resultant velocity of the aircraft. This is the vector sum of the aircraft's velocity and the wind's velocity. Since the wind is blowing towards the south, and the aircraft wants to fly due west, the resultant velocity will have a direction between west and south.

We can use the Pythagorean theorem to find the magnitude of the resultant velocity. The magnitude is equal to the square root of the sum of the squares of the two velocities:

Resultant velocity = sqrt((aircraft velocity)^2 + (wind velocity)^2)
Resultant velocity = sqrt((228 km/h)^2 + (50 km/h)^2)
Resultant velocity ≈ 236.4 km/h

So, the plane's speed relative to the ground will be approximately 236.4 km/h.

a. Vpa = Vpg + Va = -228 km/h

Vpg - 50i = -228
Vpg = -228 + 50i

Tan Ar = Y/X = 50/-228 = -0.21930
Ar = -12.4o = Reference angle.

A = -12.4 + 180 = 167.63oC CCW = 12.4o N. of W. = Heading.

b. Vpg = X/CosA = -228/Cos167.63 = 233.4
km/h. = Plane's speed relative to gnd.