In still air (no wind), a plane can fly with a speed of 150 km/h = 212 km/h. Now, a wind blows from west to east with a velocity of 150 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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BlackTom AI · Accepted Answer

First, I restate the problem as given: a plane flies in still air at some speed, and a wind blows from west to east with another speed. The question asks for the angle at which the plane must fly relative to the east–west direction so that its resultant ground track is directly north.

Note the data in the prompt is inconsistent: it states the plane can fly in still air at 150 km/h = 212 km/h, which cannot be true because 150 and 212 are different speeds for the same quantity. This signals a likely error in the numbers or units. When a problem contains such contradictory information, any conclusion drawn from it would be unreliable, so the first consideration is that the given speeds must be clarified before solving.

Even if we momentarily set aside the inconsistency and attempt to reason with a standard vector-addition model, we still face a conflict. Let the plane's airspeed (speed relative to the air) be P, and the wind speed be W, with the wind blowing east (+x). The plane can head with some angle θ relative to the east–west axis, giving it a velocity relative to the air with components: Vp_x = P cos θ, Vp_y = P sin θ (assuming θ is measured from the east toward the north). The ground velocity is the sum of the airspeed vector and the wind vector: Vg_x = Vp_x + W, Vg_y = Vp_y. To move directly north, Vg_x must be 0, and Vg_y must be positive.

Setting Vg_x = 0 gives P cos θ + W = 0, so cos θ = -W / P. Since cos θ has value in the range [-1, 1], this equation is solvable only if W ≤ P. If W > P, there is no angle θ that yields zero east–west ground component; the wind is strong enough to push the plane eastward no matter how you head.

Now apply this to the numbers as stated (even though they’re inconsistent): if P = 150 km/h and W = 150 km/h, then cos θ = -1, which gives θ = 180 degrees. That would mean heading directly west, which would make the plane’s ground track entirely west, not north, so the resulting northward component would be zero (Vg_y = P sin θ = 0). Therefore, with equal wind and plane airspeeds, you cannot achieve a purely northward ground track.

If instead the intended numbers were P > W (for example P = 212 km/h and W = 150 km/h), then cos θ = -W / P would be -150/212 ≈ -0.708, giving θ ≈ 135 degrees from the positive east axis, i.e., heading somewhat to the northwest, resulting in a northward ground track. But with the given data, the necessary condition W ≤ P is not clearly satisfied, and the math shows that a purely northward ground track cannot be achieved.

In summary, the core misconception here is assuming you can always cancel the wind’s eastward push by heading into the wind to achieve northward motion. The feasibility depends on the relative magnitudes of plane airspeed and wind speed. With the stated equal speeds (P = W), the required condition for a purely northward ground track cannot be met, so the problem as written has an inherent inconsistency and cannot be solved to yield a valid angle under those values.

If there were a complete set of answer choices, a correct solution would examine the vector condition Vg_x = 0 and check whether W ≤ P. If W > P, no angle would work; if W ≤ P, the angle would be arccos(-W/P) (measured from the east toward the north). The provided data do not satisfy those criteria in a coherent way, so any particular angle proposed (such as 45 degrees) cannot be justified as the correct solution without correcting the speeds.

Overall, the key takeaway is: ensure the wind speed does not exceed the plane’s airspeed; only then can you find a heading that yields a northward ground track, and the angle is determined by arccos(-W/P). In the present prompt, the conflicting numbers prevent a valid calculation and a unique correct angle from being determined.

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

In still air (no wind), a plane can fly with a speed of 150 km/h = 212 km/h. Now, a wind blows from west to east with a velocity of 150 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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