Relative Velocity in Two Dimensions

From a signed scalar to a vector subtraction

The one-dimensional companion set up $v_{AB} = v_A - v_B$ along a line, with sign carrying direction. In a plane the velocities point anywhere, so the subtraction must be vector-wise. NCERT's "Points to ponder" warns: the resultant velocity of an object subjected to two velocities is $\mathbf{v}_1 + \mathbf{v}_2$, but the velocity of object 1 relative to object 2 is $\mathbf{v}_{12} = \mathbf{v}_1 - \mathbf{v}_2$. Same symbols, opposite operations.

The full statement: if $\mathbf{v}_A$ and $\mathbf{v}_B$ are measured in the same ground frame, the velocity of A as seen by B is

The same rule applies to position and acceleration: $\mathbf{r}_{AB} = \mathbf{r}_A - \mathbf{r}_B$ and $\mathbf{a}_{AB} = \mathbf{a}_A - \mathbf{a}_B$. If $\mathbf{a}_A = \mathbf{a}_B$ (for example two projectiles in flight, both with $-g\,\hat{\mathbf{j}}$), then $\mathbf{a}_{AB} = \mathbf{0}$ — the relative motion of two such objects is uniform, a fact NEET exploits in projectile-collision problems.

Magnitude and direction of the relative velocity

To subtract $\mathbf{v}_B$ from $\mathbf{v}_A$, draw $\mathbf{v}_A$ and $-\mathbf{v}_B$ head-to-tail. The resultant of this addition is $\mathbf{v}_{AB}$, exactly as NCERT Fig. 3.5 illustrates for general vectors. Applying the law of cosines (NCERT Eq. 3.24a, used in Example 3.3 for the motorboat) to the triangle formed by $\mathbf{v}_A$, $\mathbf{v}_B$ and $\mathbf{v}_{AB}$:

where $\theta$ is the angle between $\mathbf{v}_A$ and $\mathbf{v}_B$ measured in the ground frame. The minus sign in front of $2 v_A v_B \cos\theta$ is what makes this a subtraction formula rather than the familiar addition $|\mathbf{R}|^2 = A^2 + B^2 + 2AB\cos\theta$. The direction of $\mathbf{v}_{AB}$ comes from the law of sines or, more cleanly, by resolving both vectors into components and subtracting:

Three special cases save mental effort:

NEET 2016 (Q.160) tests exactly the structure of this formula: if $|\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}|$, the angle between $\mathbf{A}$ and $\mathbf{B}$ must be $90°$. The relative-velocity magnitude inherits the same algebra; perpendicular vectors are the unique case where addition and subtraction give the same length.

The symmetry v_BA = −v_AB

Swap the labels: $\mathbf{v}_{BA} = \mathbf{v}_B - \mathbf{v}_A = -(\mathbf{v}_A - \mathbf{v}_B) = -\mathbf{v}_{AB}$. Two observers always see each other moving with the same speed but in opposite directions. This is more than book-keeping. If you compute $\mathbf{v}_{AB}$ and a problem then asks for $\mathbf{v}_{BA}$, no fresh calculation is needed — flip the direction, keep the magnitude. It is also a useful self-check: two consistent computations must produce vectors that are exact negatives.

River-boat-man: shortest time vs shortest path

A river flows uniformly with velocity $\mathbf{v}_r$ along its banks; a boat moves with velocity $\mathbf{v}_b$ relative to the water. The boat's ground-frame velocity is $\mathbf{v}_b + \mathbf{v}_r$. Set up axes: $\hat{\mathbf{i}}$ along the current, $\hat{\mathbf{j}}$ across the river. Let the boat be aimed at angle $\phi$ measured from $\hat{\mathbf{j}}$ (so $\phi = 0$ means straight across). Then

The cross-river component $v_b \cos\phi$ determines how long the crossing takes; the along-river component $v_r - v_b \sin\phi$ determines the drift downstream. NEET asks two sub-questions about this set-up — and they have different answers.

(a) Shortest time

Crossing time $t = d / (v_b \cos\phi)$, where $d$ is the river width. This is minimum when $\cos\phi$ is maximum, i.e. $\phi = 0$: the boat is aimed perpendicular to the banks. Then $t_\text{min} = d/v_b$, and the boat drifts downstream by $x = v_r \cdot (d/v_b)$. The boat reaches the opposite bank at the fastest possible time, but not directly opposite to where it started.

(b) Shortest path (zero drift)

For the boat to land directly opposite, the upstream component must cancel the current: $v_b \sin\phi = v_r$, so $\sin\phi = v_r / v_b$. This is only possible when $v_b > v_r$ — a boat slower than the river cannot beat the drift. The cross-river speed is then $v_b \cos\phi = \sqrt{v_b^2 - v_r^2}$ and the crossing time is $t = d / \sqrt{v_b^2 - v_r^2}$, which is longer than $d/v_b$ for any non-zero $v_r$.

Rain and man — why the umbrella tilts forward

NCERT Example 3.1 is the simpler vector-addition version: rain falls vertically, wind blows sideways, and you find the direction of the resultant rain velocity in the ground frame. The classical NEET problem is the relative-velocity version: rain falls vertically with velocity $v_r$ downward in the ground frame, and a man walks horizontally with velocity $v_m$. The velocity of rain as the man sees it is

In the man's frame the rain has a horizontal component pointing backward (against his motion) and the same downward component. The rain therefore appears to come from a direction tilted forward of vertical by an angle

measured from the vertical. To keep dry the man must tilt the umbrella through the same angle $\alpha$, forward — in the direction he is walking. This is counter-intuitive on first reading: students often imagine the rain "falling behind" them as they walk forward, and tilt the umbrella backward to "catch up". The vector picture shows the opposite. In the man's frame the rain rushes toward him from the front; the umbrella must face the front.

Aircraft in a crosswind — ground velocity

An aircraft flies with an "airspeed" of $\mathbf{v}_a$ relative to the air. The air itself moves with a wind velocity $\mathbf{v}_w$ in the ground frame. The aircraft's velocity in the ground frame — its "ground velocity" — is the vector sum $\mathbf{v}_g = \mathbf{v}_a + \mathbf{v}_w$. This is the same logic as the boat in the river, with the medium and vehicle relabelled.

NCERT Exercise 3.22 sits in the same family: an aircraft at $h = 3400$ m sweeps through $30°$ at a ground observer in $\Delta t = 10$ s. Horizontal distance $\Delta x = 2h \tan 15° \approx 1822$ m, ground speed $\approx 182$ m s$^{-1}$. The geometry is set up in the ground frame — and is what feeds any later relative-velocity calculation between this aircraft and another.

The closest-approach trick — fix one observer

For two objects A and B moving with arbitrary velocities, finding the moment and distance of closest approach reduces to one dimension by sitting in B's frame. There B is stationary and A moves with $\mathbf{v}_{AB} = \mathbf{v}_A - \mathbf{v}_B$. The closest approach is the perpendicular distance from B to the straight line along which A moves — pure geometry. It is the engine behind every "minimum distance between two ships" or "do these two cars collide" question. NCERT Exercise 3.20 reminds you why it is legitimate: $\mathbf{v}_{\text{avg}} = [\mathbf{r}(t_2) - \mathbf{r}(t_1)] / (t_2 - t_1)$ holds for arbitrary motion, so the kinematic identities transfer cleanly into the relative frame.