In still air (no wind), a plane can fly with a speed of 150
√
2
km/h = 212 km/h. Now, a wind blows from west to east with a velocity of 150
√
2
km/h=212km/h. At what angle compared to the east-west direction does the plane need to fly to move directly north?

Question

In still air (no wind), a plane can fly with a speed of 150
√
2
 km/h = 212 km/h. Now, a wind blows from west to east with a velocity of 150
√
2
 km/h=212km/h. At what angle compared to the east-west direction does the plane need to fly to move directly north?

BlackTom AI · Accepted Answer

We start by laying out the given vectors and the condition for moving directly north.
- Airspeed (plane relative to air): 212 km/h at some heading angle θ measured from east toward north, so its components are: x-component = 212 cosθ (east positive), y-component = 212 sinθ (north positive).
- Wind: 212 km/h from west to east, so it adds a vector with components: x = +212 (east), y = 0.
- Ground velocity is the sum of airspeed and wind: V_ground = (212 cosθ + 212, 212 sinθ).
To move directly north, the east-west (x) component of the ground velocity must be zero:
212 cosθ + 212 = 0 → cosθ = -1.
But cosθ = -1 implies θ = 180°, which corresponds to the plane pointing directly west relative to the air. At θ = 180°, the north component is 212 sin(180°) = 0, so the ground velocity would be (0, 0), not north.
This reveals a crucial point: with wind speed equal to the plane’s airspeed and blowing exactly eastward, there is no heading that yields a nonzero north component while canceling the east component to zero. In other words, you cannot move directly north under these conditions because cancelling the eastward wind leaves no northward component.

Now evaluating a concrete candidate option, such as 45°:
- If θ = 45°, the airspeed components are: x = 212 cos45° ≈ 212 × 0.707 ≈ 150 (east), y = 212 sin45° ≈ 150 (north).
- Adding wind: ground x = 150 + 212 = 362 (east), ground y = 150 (north).
This results in a ground trajectory that is steeply to the east, not north, confirming that 45° does not satisfy the requirement.

Therefore, the proposed heading of 45° does not achieve a direct northward track, and in fact, under the given equal magnitudes and opposing directions, there is no heading that yields pure northward motion.

Section Merge: PHYS 211: Mechanics (UP; Spring 2025) phys211_MT2

In still air (no wind), a plane can fly with a speed of 150 √ 2 km/h = 212 km/h. Now, a wind blows from west to east with a velocity of 150 √ 2 km/h=212km/h. At what angle compared to the east-west direction does the plane need to fly to move directly north?

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