Motion is not absolute — it is always observed from a reference frame. Understanding how velocities transform between frames is the foundation of classical mechanics.
A reference frame is a coordinate system attached to an observer. There is no single "correct" frame — all inertial frames (non-accelerating) are equivalent. When we say an object has velocity $v$, we always mean velocity relative to some frame.
The key insight: the same physical event can have completely different descriptions depending on who is measuring it. A passenger sitting still in a train is moving at 100 km/h relative to the platform. Both descriptions are equally valid.
$S$inertial reference frame
$S'$second (moving) frame
$\vec{v}_{A/B}$velocity of A relative to Bm s⁻¹
$\vec{u}$velocity of frame itselfm s⁻¹
🚂
Train frame
A ball rolled forward at 5 m/s inside a train moving at 30 m/s appears to move at 35 m/s to a stationary observer.
⚡
Inertial = non-accelerating
Galilean addition applies only to inertial frames. An accelerating frame introduces fictitious forces (Coriolis, centrifugal).
🔁
Antisymmetry
$\vec{v}_{A/B} = -\vec{v}_{B/A}$. If you move at +20 m/s relative to me, I move at −20 m/s relative to you.
§ 02Core Rule
Galilean velocity addition
The rule for combining velocities in classical mechanics is simple vector addition. The subscript notation is crucial — learn to read it fluently.
GALILEAN ADDITION LAW
$$\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$$
Read: "velocity of A relative to C equals velocity of A relative to B, plus velocity of B relative to C."
The middle subscripts $B$ cancel, like fractions: $A/B \cdot B/C = A/C$.
Derivation from position
$\vec{r}_{A/C} = \vec{r}_{A/B} + \vec{r}_{B/C}$
Position of A measured from C = position of A from B, plus position of B from C (vector triangle)
By definition of velocity as $d\vec{r}/dt$. This holds exactly for $v \ll c$; near light speed, use Lorentz addition.
Subscript trick. Write the subscript chain so adjacent letters match and cancel: $\vec{v}_{P/Q} + \vec{v}_{Q/R} = \vec{v}_{P/R}$. This works for any length chain: $\vec{v}_{A/B} + \vec{v}_{B/C} + \vec{v}_{C/D} = \vec{v}_{A/D}$.
Acceleration transforms the same way between inertial frames: $\vec{a}_{A/C} = \vec{a}_{A/B} + \vec{a}_{B/C}$. Since a constant-velocity frame has $\vec{a}_{B/C} = 0$, the acceleration of an object looks the same in all inertial frames — this is why Newton's laws apply equally in all of them.
§ 03One Dimension
1D relative velocity
In one dimension, vectors reduce to signed scalars. The addition law becomes simple algebra — but direction (sign) still matters critically.
Car A: $v_{A/G} = +30$ m/s
Car B: $v_{B/G} = +20$ m/s
$v_{A/B} = v_{A/G} + v_{G/B} = 30 + (-20) = \mathbf{+10}$ m/s A gains on B slowly from behind.
Opposite directions
Car A: $v_{A/G} = +30$ m/s
Car B: $v_{B/G} = -25$ m/s
$v_{A/B} = 30 - (-25) = \mathbf{+55}$ m/s They close at combined speed.
Overtaking
Train: $v_{T/G} = +50$ m/s
Passenger walks: $v_{P/T} = +1.5$ m/s
$v_{P/G} = 1.5 + 50 = \mathbf{+51.5}$ m/s Walking forward on a train adds to ground speed.
Conveyor belt
Belt: $v_{B/G} = +2$ m/s
Walker on belt: $v_{W/B} = -2$ m/s
$v_{W/G} = -2 + 2 = \mathbf{0}$ m/s Walking backward at the belt speed: stationary to ground.
Time to meet / overtake
$$t_{\text{meet}} = \frac{d}{|v_{A/B}|}$$
Time for A to travel distance $d$ relative to B. Use the relative speed, not individual speeds.
$$d_{\text{at }t} = v_{\text{rel}} \cdot t$$
Separation changes at the relative speed. Closing = positive relative approach velocity.
Exam tip: to find when two objects meet, work in one object's reference frame. In that frame, one object is stationary — the problem reduces to simple distance ÷ speed.
§ 04Two Dimensions
2D relative velocity
In 2D, the addition law is identical — but now the velocities are vectors with $x$ and $y$ components. Add components separately, then find the resultant magnitude and direction.
Check quadrant — $\arctan$ only gives $[-90°, 90°]$. Use $\text{atan2}(v_y, v_x)$ for full range.
General 2D strategy
Identify all frames in the problem and label them clearly (e.g. Ground $G$, Wind $W$, Aircraft $A$).
Naming frames prevents subscript confusion
Write the velocity chain: $\vec{v}_{A/G} = \vec{v}_{A/W} + \vec{v}_{W/G}$
Middle subscripts cancel; left side is what you want to find
Resolve each known vector into $(x, y)$ components using $\cos$ and $\sin$.
Work in Cartesian coordinates — avoid trying to "see" the geometry directly
Add $x$ components and add $y$ components separately.
The axes are independent
Reconstruct $|\vec{v}_{A/G}|$ and direction $\theta$ from the components.
Pythagoras + $\arctan$, mindful of quadrant
§ 05Classic Problem
River crossing problem
A boat crosses a river of width $d$. The river has current speed $v_c$ (parallel to banks). The boat's engine gives it speed $v_b$ relative to the water. The direction the boat points (heading angle $\alpha$ from the bank) determines the actual path over the ground.
$y$ = across-river. Only the boat's engine provides crossing speed.
Three objectives — three solutions
Objective
Required heading $\alpha$
Result
Cross in shortest time
$\alpha = 90°$ (point straight across)
$t_{\min} = d/v_b$ but drifts $\Delta x = v_c \cdot d/v_b$
Land directly opposite
$\alpha = 90° + \arcsin(v_c/v_b)$ (aim upstream)
Zero drift, longer time $t = d/(v_b\sin\alpha)$. Requires $v_b > v_c$.
Cross in shortest distance
$\alpha = 90° + \arcsin(v_c/v_b)$ (same as above)
Straight-line path. The shortest path is always the zero-drift path.
$v_b < v_c$ (can't go straight)
Aim as far upstream as possible: $\alpha$ maximising $v_y/v_x$
Minimise drift; cannot reach directly opposite. Some downstream drift unavoidable.
Critical condition: zero drift requires $\sin\alpha_{\text{upstream}} = v_c / v_b$. This only has a solution if $v_c \leq v_b$. If the current is faster than the boat, you cannot cross straight — you will always be swept downstream.
MINIMUM TIME CROSSING (α = 90°)
$$t = \frac{d}{v_b}$$
Time depends only on crossing component. Current has no effect on crossing time when heading is perpendicular.
$$\text{drift} = v_c \cdot t = \frac{v_c \cdot d}{v_b}$$
Downstream displacement. Increases with current speed and width.
§ 06Apparent Velocity
Rain, wind, and apparent motion
When an observer is moving, stationary objects appear to move. Raindrops falling vertically appear to fall at an angle to a person who is running. This apparent velocity is simply relative velocity.
To keep dry: tilt umbrella forward at angle $\phi$ from vertical. Faster walking → more tilt needed.
Wind correction for aircraft. An aircraft wants to fly on heading $\theta_{\text{desired}}$. Wind blows at $\vec{v}_w$. The pilot must aim the nose at $\theta_{\text{aim}} = \theta_{\text{desired}} - \arcsin(v_w / v_{\text{air}})$ (adjusted for wind direction). The ground track is what matters; the heading is what the pilot controls.
$\vec{v}_{\text{ground}} = \vec{v}_{\text{air}} + \vec{v}_{\text{wind}}$
To fly north in east wind: point nose NW.
Correction angle: $\arcsin(v_w / v_a)$.
Groundspeed $\neq$ airspeed in crosswind.
Relative motion of ships
Ship A and B have velocities $\vec{v}_A, \vec{v}_B$.
B's motion relative to A: $\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A$.
Collision course: $\vec{v}_{B/A}$ points toward A's current position.
Swimmer in current
Identical to river crossing.
Swimmer aims at angle to current to compensate.
Effective crossing speed: $\sqrt{v_s^2 - v_c^2}$ when aiming to land directly opposite.
§ 07Interactive
Live simulations
Four scenarios — each demonstrates a different aspect of relative velocity. Adjust sliders and toggle the reference frame to see the same motion described differently.