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Irodov 1.6Trickykinematicsrelative-motionvectors

Wind Velocity Relative to a Ship

Problem 1.6
A ship moves along the equator to the east with velocity v0=30 km/hv_0 = 30 \text{ km/h}. The southeastern wind blows at an angle φ=60∘\varphi = 60^\circ to the equator with velocity v=15 km/hv = 15 \text{ km/h}. Find the wind velocity v′v' relative to the ship and the angle φ′\varphi' between the equator and the wind direction in the reference frame fixed to the ship.
Answer: v′≈40 km/h,  φ′≈19∘v' \approx 40 \text{ km/h},\; \varphi' \approx 19^\circ
Solution Path
A ship moving east at 30 km/h experiences a 15 km/h southeastern wind as a stronger, nearly head-on wind. By vector subtraction, the relative wind speed is about 40 km/h at just 19 degrees to the equator. The ship's motion flattens the wind angle and increases the apparent speed.
01Problem Restatement
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A ship is sailing eastward along the equator at a steady speed. A wind is blowing from the southeast, making an angle with the equator. We want to find what the wind looks like from the ship's perspective.Given:
- v0=30v_0 = 30 km/h (ship speed, directed east along the equator)
- v=15v = 15 km/h (wind speed, blowing from the southeast)
- φ=60∘\varphi = 60^\circ (angle between the wind direction and the equator)We need to find: the wind velocity v′v' relative to the ship (in km/h) and the angle φ′\varphi' between the apparent wind direction and the equator (in degrees).
Find v′v' and φ′\varphi' in the ship's reference frame
02Physical Picture and Strategy
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Picture a cargo ship steaming east along the equator on a calm sea. A wind is blowing from the southeast, hitting the ship at an angle. A sailor standing on the deck feels a different wind than a person standing on shore would feel, because the ship's own motion adds to the apparent wind.We set up a coordinate system with the x-axis pointing east and the y-axis pointing north.The ship's velocity is purely eastward:v⃗0=(30,  0) km/h\vec{v}_0 = (30,\; 0) \text{ km/h}The wind blows FROM the southeast. "Southeastern wind" means the wind originates in the southeast, so it blows toward the northwest. The velocity vector points in the NW direction, at 60∘60^\circ above the negative x-axis:v⃗wind=(−vcos⁡60∘,  +vsin⁡60∘)=(−7.5,  12.99) km/h\vec{v}_{\text{wind}} = (-v\cos 60^\circ,\; +v\sin 60^\circ) = (-7.5,\; 12.99) \text{ km/h}Strategy: We compute the relative velocity v⃗′=v⃗wind−v⃗ship\vec{v}' = \vec{v}_{\text{wind}} - \vec{v}_{\text{ship}} by vector subtraction, then find its magnitude and direction angle.
v⃗wind=(−7.5,  12.99)\vec{v}_{\text{wind}} = (-7.5,\; 12.99) km/h
03Step 1: Compute the Relative Wind Velocity
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To find the wind as experienced on the ship, we subtract the ship's velocity from the wind's velocity. This is the standard relative velocity formula: the velocity of the wind in the ship's frame equals the wind velocity in the ground frame minus the ship's velocity.Using the relative velocity formula:v⃗′=v⃗wind−v⃗ship\vec{v}' = \vec{v}_{\text{wind}} - \vec{v}_{\text{ship}}Computing each component:vx′=−7.5−30=−37.5 km/hv'_x = -7.5 - 30 = -37.5 \text{ km/h}vy′=12.99−0=12.99 km/hv'_y = 12.99 - 0 = 12.99 \text{ km/h}v⃗′=(−37.5,  12.99) km/h\vec{v}' = (-37.5,\; 12.99) \text{ km/h}This means: the apparent wind has a large westward component (−37.5-37.5) and a smaller northward component (12.9912.99). The ship's eastward motion makes the wind appear to blow much more strongly from ahead (the west).
v⃗′=(−37.5,  12.99)\vec{v}' = (-37.5,\; 12.99) km/h
04Step 2: Why the Wind Shifts DirectionKEY INSIGHT
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Compare what happens in the two frames of reference.Ground frame: A person standing on shore sees the wind blowing from the SE at 60∘60^\circ to the equator at 1515 km/h. The wind has a modest westward component and a large northward component.Ship frame: The sailor on the moving ship feels a much stronger apparent wind. Because the ship moves east at 3030 km/h, the sailor experiences an additional 3030 km/h headwind from the east. This large extra westward component swamps the original wind's westward component.The result: the apparent wind is much stronger (≈40\approx 40 km/h vs 1515 km/h) and appears to blow almost directly from the west, with the angle to the equator dropping from 60∘60^\circ to just 19∘19^\circ.This means: the ship's motion "flattens" the wind direction toward the equator and dramatically increases the apparent wind speed. This is exactly the effect sailors experience in practice: moving into the wind makes it feel stronger and more head-on.
Ship's motion flattens wind angle from 60∘60^\circ to 19∘19^\circ
05Step 3: Calculate Magnitude and Angle
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Now we compute the magnitude of v⃗′\vec{v}' using the Pythagorean theorem and the angle using arctangent.Magnitude (using ∣v⃗′∣=vx′2+vy′2|\vec{v}'| = \sqrt{v'^2_x + v'^2_y}):∣v⃗′∣=(−37.5)2+(12.99)2|\vec{v}'| = \sqrt{(-37.5)^2 + (12.99)^2}=1406.25+168.74=1575= \sqrt{1406.25 + 168.74} = \sqrt{1575}v′≈39.7 km/h\boxed{v' \approx 39.7 \text{ km/h}}Angle (using φ′=arctan⁡(∣vy′∣/∣vx′∣)\varphi' = \arctan(|v'_y| / |v'_x|)):φ′=arctan⁡ ⁣(12.9937.5)=arctan⁡(0.3464)\varphi' = \arctan\!\left(\frac{12.99}{37.5}\right) = \arctan(0.3464)φ′≈19.1∘\boxed{\varphi' \approx 19.1^\circ}This means: the sailor feels a wind of about 4040 km/h blowing at a shallow 19∘19^\circ angle above the equator, nearly a headwind.
v′≈40v' \approx 40 km/h, φ′≈19∘\varphi' \approx 19^\circ
06Verification and Summary
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Verification:Dimensional check: We subtracted two velocities (km/h) to get a velocity (km/h), then used Pythagorean theorem on velocity components. All dimensions are consistent. ✓\checkmarkReasonableness: The apparent wind speed (4040 km/h) is larger than both the actual wind speed (1515 km/h) and the ship speed (3030 km/h). This makes sense because the ship moves into the wind's westward component, adding to it. ✓\checkmarkAngle check: The angle φ′=19∘<φ=60∘\varphi' = 19^\circ < \varphi = 60^\circ. The ship's eastward motion adds a large westward component to the apparent wind, making the wind appear to come from nearly due west. The angle to the equator must decrease. ✓\checkmarkLimiting case: If the ship were stationary (v0=0v_0 = 0), the apparent wind would equal the actual wind: v′=15v' = 15 km/h at 60∘60^\circ. Our formula gives v⃗′=v⃗wind−0⃗=v⃗wind\vec{v}' = \vec{v}_{\text{wind}} - \vec{0} = \vec{v}_{\text{wind}}. ✓\checkmarkSummary: A ship sailing east at 3030 km/h experiences a southeastern wind (1515 km/h, 60∘60^\circ) as a much stronger, nearly head-on wind. The relative wind speed is about 4040 km/h, and it appears to blow at only 19∘19^\circ to the equator. The key was recognizing that "southeastern wind" means the velocity vector points northwest, and then applying vector subtraction to switch to the ship's frame.
v′≈40 km/h,φ′≈19∘\boxed{v' \approx 40 \text{ km/h}, \quad \varphi' \approx 19^\circ}
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