$v'=\sqrt{v_0^2+v^2+2v_0v\cos\varphi}\approx39.7\text{ km/h}$, at $\varphi'\approx19.1^\circ$ to the equator.
(ii) Problem Statement
"A ship moves along the equator to the east with velocity $v_0=30$ km/hour. The southeastern wind blows at an angle $\varphi=60^\circ$ to the equator with velocity $v=15$ km/hour. Find the wind velocity $v'$ relative to the ship and the angle $\varphi'$ between the equator and the wind direction in the reference frame fixed to the ship."
Reference Answer Verification (from the supplied solution image):
The reference writes $\vec v'=\vec v-\vec v_0$ and uses the triangle to get $v'=\sqrt{v_0^2+v^2+2v_0v\cos\varphi}=39.7$ km/h. The sine rule $\dfrac{v'}{\sin(\pi-\varphi)}=\dfrac{v}{\sin\theta}$ gives $\theta=\sin^{-1}\!\left(\dfrac{v\sin\varphi}{v'}\right)=19.1^\circ$. Our work matches.
Reading the words: a south-eastern wind comes from the south-east, so it blows toward the north-west. The angle $\varphi$ is measured between that wind direction and the equator, as in the figure.
(iii) The Big Idea in Plain Words
Speed depends on who is watching. A person on a moving ship feels a different wind from a person on the shore.
Symbols: $\vec v_0$ is the ship's velocity. $\vec v$ is the wind's velocity. $\vec v'$ is the wind's velocity as the ship's crew sees it. An arrow over a letter means it has a size and a direction. Here east is the $x$ direction and north is the $y$ direction.
The crew feels the wind relative to themselves. So we subtract the ship's velocity from the wind's velocity: $\vec v'=\vec v-\vec v_0$. Then we find the length of $\vec v'$ and its angle.
(iv) The Solution Viewed in Each Reference Frame
1. Ground or Lab Frame (Observer on the Shore)
Origin fixed on the equatorWhy: first write both velocities as east and north parts (components).
$\vec v_0=(30,\ 0)$ km/h. The wind blows toward the north-west, so $\vec v=(-v\cos\varphi,\ v\sin\varphi)=(-7.5,\ 13.0)$ km/h.
Why: the wind relative to the ship is the wind velocity minus the ship velocity.
$$\vec v'=\vec v-\vec v_0=(-37.5,\ 13.0)\text{ km/h}$$
Why: the size of an arrow comes from its two parts, using Pythagoras.
$$v'=\sqrt{37.5^2+13.0^2}=\sqrt{1575}\approx39.7\text{ km/h}$$
Why: the angle with the equator comes from the tangent of "rise over run".
$$\tan\varphi'=\frac{13.0}{37.5}=0.346\;\Rightarrow\;\varphi'\approx19.1^\circ$$
The crew feels the wind blowing toward the west-north-west, $19.1^\circ$ north of west.
2. Ship Frame (The Frame the Question Asks For)
Origin moves with the shipWhy: in this frame the ship is at rest. Subtract $\vec v_0$ from every velocity.
Ship: $\vec v_0-\vec v_0=0$. The shore moves at $-\vec v_0=(-30,\ 0)$ km/h.
Why: the wind's velocity in this frame is exactly what the crew measures.
$\vec v'=\vec v-\vec v_0=(-37.5,\ 13.0)$ km/h, so $v'\approx39.7$ km/h and $\varphi'\approx19.1^\circ$. Nothing more is needed.
3. Medium Frame (Observer Drifting With the Air)
Origin moves at $\vec v$Why: the medium here is the air. Riding with it, the air is at rest, so the wind is $0$.
The ship now moves with velocity $\vec v_0-\vec v=(37.5,\ -13.0)$ km/h. That equals $-\vec v'$.
Why: the crew sees the wind blowing against their motion through the still air.
So the wind relative to the ship is $-(\vec v_0-\vec v)=\vec v'$. The size is $39.7$ km/h. The direction is $19.1^\circ$ north of west. The ship moves $19.1^\circ$ south of east in the air. Only relative speeds matter, and the answer is the same.
4. Centre-of-Mass (CM) Frame
This frame does not apply. The air is not a single object with a known mass, and the ship and air do not form a closed two-body system. Relative velocity needs no centre of mass.
What Changes and What Stays the Same Across Frames
- What changes: the ship's velocity and the wind's own velocity, which depend on who is watching.
- What stays the same: the wind relative to the ship, $\vec v'=\vec v-\vec v_0$. Its size $39.7$ km/h and its angle $19.1^\circ$ are identical in every frame.
(v) The Solution by Each Method
Method 1: Relative-Motion Method (Components)
Fastest- Why: riding on the ship, the ship stands still and the wind velocity is $\vec v-\vec v_0$.
$\vec v'=(-v\cos\varphi-v_0,\ v\sin\varphi)=(-37.5,\ 13.0)$ km/h. - Why: length and angle come from the two parts.
$v'=\sqrt{1575}=39.7$ km/h and $\varphi'=\tan^{-1}\dfrac{13.0}{37.5}=19.1^\circ$.
Method 2: Graph Method (Velocity Triangle)
- Why: draw $\vec v_0$ and $\vec v$ from one point. Then $\vec v'$ joins the tip of $\vec v_0$ to the tip of $\vec v$.
The angle between $\vec v$ and the equator is $\varphi$. So the angle between $\vec v$ and $\vec v_0$ is $\pi-\varphi$. - Why: the cosine rule gives the third side of a triangle.
$v'^2=v_0^2+v^2-2v_0v\cos(\pi-\varphi)=v_0^2+v^2+2v_0v\cos\varphi$, so $v'=39.7$ km/h. - Why: the sine rule gives the angle.
$\dfrac{v'}{\sin(\pi-\varphi)}=\dfrac{v}{\sin\varphi'}\Rightarrow\sin\varphi'=\dfrac{v\sin\varphi}{v'}=0.327$, so $\varphi'=19.1^\circ$.
Method 3: Calculus Method
- Why: follow one bit of wind. Its position relative to the ship is its position minus the ship's position.
$\vec r'(t)=\vec r_{wind}(t)-\vec r_{ship}(t)$, with $\vec r_{wind}=\vec r_0+\vec vt$ and $\vec r_{ship}=\vec v_0t$. - Why: velocity is the rate of change of position, so differentiate.
$\vec v'=\dfrac{d\vec r'}{dt}=\vec v-\vec v_0=(-37.5,\ 13.0)$ km/h. - Why: read off the size and angle as before.
$v'=39.7$ km/h and $\varphi'=19.1^\circ$.
Methods 4 and 5: Force and Energy Methods
Force Method: not suited. All velocities are constant, so no net force acts.
Energy Method: not suited. No mass or work is given.
Fastest: Method 1. Most reliable in an exam: Method 1 with components, since it works for any angle $\varphi$ and avoids the sign slip in the cosine rule. Method 2 is a good check. All give $v'\approx39.7$ km/h and $\varphi'\approx19.1^\circ$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|---|---|---|
| $v_0$ | ship speed, toward the east | km/h | |
| $v$ | wind speed (the wind blows toward the north-west) | km/h | |
| $\varphi$ | angle between the wind and the equator (ground frame) | degrees | |
| $v'$ | wind speed relative to the ship (answer) | km/h | |
| $\varphi'$ | angle between that wind and the equator (answer) | degrees | |
| $\vec v'$ | components (east, north) | km/h |
(vii) Common Exam Traps
- Adding the velocities instead of subtracting. The wind relative to the ship is $\vec v-\vec v_0$.
- Writing $-2v_0v\cos\varphi$ in the cosine rule. The angle between $\vec v$ and $\vec v_0$ is $\pi-\varphi$, which gives a plus sign.
- Pointing the wind the wrong way. A south-eastern wind blows toward the north-west.
- Assuming $\varphi'=\varphi$. The ship's motion changes the angle, from $60^\circ$ to $19.1^\circ$.
- Using $\sin^{-1}$ only. It gives angles up to $90^\circ$. Use components ($\tan^{-1}$ with signs) for larger angles.
- Mixing units. Keep everything in km/h.
Verified: the live numbers, all frames and all methods give $v'\approx39.7$ km/h and $\varphi'\approx19.1^\circ$ by default, with units in km/h and degrees.