A pilot must fly a plane due north to reach the destination. The plane has a still air speed of 300 km/h. A northeast wind blows at 90 km/h. (a) What is the speed of the plane relative to the ground? (b) In which direction must the pilot steer the plane to fly due north?

(a) To find the speed of the plane relative to the ground, we need to consider both the still air speed of the plane and the wind speed and direction. Since the wind is blowing from the northeast, it has a component to the south and a component to the west. This means the wind will affect the plane's motion in both the east-west and north-south directions.

To solve this problem, we can use vector addition. Draw a vector to represent the plane's still air speed pointing north (300 km/h), and a vector to represent the wind speed (90 km/h) pointing to the southwest (since the wind comes from the northeast).

To determine the effective speed of the plane relative to the ground, we need to subtract the southward component of the wind from the plane's still airspeed and add the westward component of the wind to the plane's velocity. Remember that the northeast wind means it's blowing towards the southwest.

Since the wind speed is 90 km/h in a northeast direction, it has equal components towards the south and towards the west, due to the 45-degree angle. Each component can be found using trigonometric functions or recognizing that in an isosceles right triangle (45-45-90), the legs are equal. So, the wind has a 90 km/h / √2 ≈ 63.64 km/h component to the south and a 63.64 km/h component to the west.

The ground speed of the plane (its speed over the ground) would be the vector sum of the plane's airspeed and the wind's effect. The northward component of the plane's velocity is reduced by the wind's southward component:

Northward velocity = 300 km/h - 63.64 km/h (southward component of the wind speed) ≈ 236.36 km/h

The westward (perpendicular) component is simply the westward component of the wind's speed, since the plane's still air speed has no east-west component:

Westward velocity = 63.64 km/h

So the speed of the plane relative to the ground (the magnitude of the ground velocity vector) can be found using the Pythagorean theorem:

Ground speed = √(Northward velocity^2 + Westward velocity^2) ≈ √(236.36^2 + 63.64^2) ≈ √(55882.37 + 4051.09) ≈ √(59933.46) ≈ 244.81 km/h

Thus, the ground speed of the plane is approximately 244.81 km/h.

(b) To determine the direction in which the pilot must steer the plane to fly due north, we need to find the angle that the plane's direction must offset the wind's westward push.

We can use trigonometry to find this angle. Specifically, we can use the tangent function, which is the ratio of the opposite side to the adjacent side in a right-angled triangle. The angle θ is the angle the pilot must steer east of due north to counteract the westward wind component:

tan(θ) = Westward velocity / Northward velocity tan(θ) = 63.64 / 236.36 θ = arctan(63.64 / 236.36) θ ≈ arctan(0.269) (using a calculator) θ ≈ 15.1° (using a calculator)

Therefore, the pilot must steer the plane approximately 15.1 degrees east of due north to account for the wind and still travel directly north.