Definition and the sign convention you must fix first
NCERT §2.5 introduces relative velocity through a chain of three observations. First, "velocity of a body is always specified with respect to some other body" (NIOS §2.1.2 puts it identically). Second, if two bodies $A$ and $B$ move along the same straight line with velocities $v_A$ and $v_B$ measured by a ground observer, the rate at which $B$'s position changes as seen from $A$ is:
$$v_{BA} = v_B - v_A \qquad \text{(velocity of } B \text{ as seen from } A\text{)}$$Third, the subscripts encode direction: $v_{BA}$ means "velocity of $B$ with respect to $A$" — $A$ is the observer, $B$ is the observed. Reversing the order reverses the sign:
$$v_{AB} = v_A - v_B = -\,v_{BA}$$In one dimension, the vector arrows of NCERT's general formula reduce to plain signed numbers. This is the freedom 1-D gives you — but it is also the trap, because students stop watching signs the moment vectors disappear. Before substituting anything, do this:
- Choose a positive direction — for example, "rightward is positive" or "from south to north is positive". Write it down at the top of your scratch sheet.
- Attach a sign to every velocity the problem gives you, based on its direction relative to your chosen axis. A body moving leftward at $40$ km/h becomes $v = -40$ km/h.
- Plug signed numbers into $v_{BA} = v_B - v_A$ without further fuss. The output sign tells you the direction of the relative velocity.
This discipline matters because NCERT's Example (NIOS §2.1.2) does exactly that. Train $A$ moves north-to-south at $60$ km/h, train $B$ moves south-to-north at $70$ km/h. Taking south-to-north as positive: $v_B = +70$, $v_A = -60$. Then $v_{BA} = +70 - (-60) = +130$ km/h. The positive sign confirms $B$ moves away from $A$ in the chosen positive direction; the magnitude $130$ km/h is the rate at which the trains close on (or pull away from) each other.
Same direction vs opposite direction — one formula, two faces
Once signs are in place, the same equation $v_{BA} = v_B - v_A$ produces two distinct outcomes depending on the geometry. The table below contrasts them with the canonical NCERT-style numbers — two cars on the same road, then two trains on parallel tracks.
| Geometry | Signed velocities (positive = same as motion of $A$) | $v_{BA} = v_B - v_A$ | Physical meaning |
|---|---|---|---|
| Same direction. $A$ chases $B$: car $A$ at $60$ km/h, car $B$ at $40$ km/h ahead | $v_A = +60$, $v_B = +40$ | $+40 - (+60) = -20$ km/h | From $A$'s frame, $B$ recedes backward at $20$ km/h, i.e. $A$ closes the gap at $20$ km/h. Closing (or overtaking) speed is the magnitude. |
| Same direction. Both at identical speed — convoy | $v_A = +50$, $v_B = +50$ | $+50 - 50 = 0$ | Relative velocity vanishes; bodies look stationary to each other even though both are moving. |
| Opposite directions. Two trains on parallel tracks; positive = north | $v_A = -60$ (going south), $v_B = +70$ (going north) | $+70 - (-60) = +130$ km/h | From $A$'s frame, $B$ approaches and then recedes at $130$ km/h. Magnitudes add because signs differ. |
| Opposite directions, equal magnitudes — head-on closure | $v_A = -25$, $v_B = +25$ | $+25 - (-25) = +50$ m/s | Closing speed is twice the individual speed. This is why head-on collisions are catastrophic at "highway" speeds. |
| Same direction; chaser slower than the led — gap widens | $v_A = +40$, $v_B = +60$ (with $B$ ahead) | $+60 - (+40) = +20$ km/h | $B$ pulls away from $A$ at $20$ km/h. $A$ never catches $B$. Sign of $v_{BA}$ tells you which one is gaining. |
Two patterns are worth committing to memory. First, when the bodies move in the same direction, the magnitude of the relative velocity is the difference of speeds — small and intuitive ("the slow car barely creeps past the slower car"). Second, when they move in opposite directions, the magnitude is the sum of speeds — much larger and the reason oncoming traffic looks frightening at highway speeds. Both are special cases of the same signed subtraction; the difference is only in the signs that go in.
Why $v_{BA} = -v_{AB}$, and why the subscript order matters
Taking the algebraic identity at face value:
$$v_{AB} = v_A - v_B, \qquad v_{BA} = v_B - v_A = -(v_A - v_B) = -v_{AB}$$So the two relative velocities have the same magnitude and opposite signs. Physically: if you (at $A$) see $B$ moving rightward at $20$ km/h, then $B$ sees you moving leftward at $20$ km/h. The relative geometry between the two frames is symmetric; only the direction reverses depending on whose seat you occupy.
NEET examiners exploit this with subscript-order traps. A question that asks "velocity of $A$ with respect to $B$" is asking for $v_{AB}$; "velocity of $B$ with respect to $A$" asks for $v_{BA}$. They differ by a minus sign. Read the question twice before you compute; the rule of thumb is — "the body whose name appears first in the subscript is the body being observed; the second is the observer".
Relative acceleration in one dimension
Acceleration is the rate of change of velocity, and exactly the same subtraction logic carries over:
$$a_{BA} = a_B - a_A$$In one dimension, with both accelerations signed, $a_{BA}$ is the acceleration of $B$ as measured from a frame moving with $A$. For NEET kinematics in Class 11, two situations dominate. First, when both bodies have the same acceleration — for example, two stones in free fall with $a_A = a_B = -g$ — the relative acceleration is zero, and in the relative frame the bodies behave as if there is no gravity. This is why two freely falling stones, released at any instants, maintain a relative velocity that changes only because of the initial velocity difference and never because of $g$.
Second, when one body accelerates and the other moves uniformly, the relative motion in the chasing frame is itself uniformly accelerated. The kinematic equations of the three-equation set apply directly in the relative frame, with $v_{rel}$, $a_{rel}$ and $s_{rel}$ replacing the ground-frame quantities. NIOS Example 2.9, where car $B$ at $70$ km/h decelerating at $20$ km/h² catches up with car $A$ cruising at $60$ km/h, is solved exactly this way.
Time-to-meet and time-to-overtake — the one-equation shortcut
The single biggest practical payoff of relative velocity is collapsing a two-body problem into a one-body problem. Without it, you write two position-versus-time equations and solve simultaneously. With it, you sit in one body's frame, treat that body as stationary, and watch the other approach you with the relative velocity. The time becomes:
$$t = \frac{\text{initial separation}}{|v_{rel}|}$$For overtaking on a straight road, "initial separation" is the gap between the chaser and the leader. For two trains crossing each other, it is the sum of their lengths — because crossing is complete only when the tail of one passes the head of the other. For head-on meeting, it is the gap between the two bodies. In every case, the algebra reduces to one division.
The cleanness of this method, in one paragraph
Consider car $A$ at $v_A = +60$ km/h chasing car $B$ at $v_B = +40$ km/h, with $B$ initially $200$ m ahead. The relative velocity of $A$ with respect to $B$ is $v_{AB} = v_A - v_B = +20$ km/h. Convert to m/s: $20$ km/h $= 20 \times \dfrac{1000}{3600} \approx 5.56$ m/s. Time to overtake: $t = 200 / 5.56 \approx 36$ s. No simultaneous equations, no quadratics — just one division. The price you pay for the shortcut is the discipline of carrying signs correctly, which is exactly what the trap callouts above were designed to enforce.
Worked examples
Car $A$ moves along a straight highway at $60$ km/h. Car $B$ moves on the same highway, in the same direction, at $40$ km/h, with $B$ exactly $200$ m ahead of $A$ at $t = 0$. How long does $A$ take to overtake $B$, and what is the closing speed?
Take rightward (direction of motion) as positive: $v_A = +60$ km/h, $v_B = +40$ km/h. Velocity of $A$ as seen from $B$:
$$v_{AB} = v_A - v_B = +60 - (+40) = +20 \text{ km/h}$$
Positive sign means $A$ approaches $B$ from behind. Convert to SI: $v_{AB} = 20 \times \dfrac{5}{18} = 5.\overline{5}$ m/s.
In $B$'s frame, $B$ is at rest and $A$ approaches with this relative speed. Time to close a gap of $200$ m:
$$t = \dfrac{200 \text{ m}}{5.56 \text{ m/s}} \approx 36 \text{ s}$$
Closing (or overtaking) speed is $20$ km/h. Note: the ground-frame speeds $60$ and $40$ km/h play no further role once $v_{AB}$ is in hand — the entire two-car problem reduces to "one car closing a gap at $20$ km/h".
Two trains, each $200$ m long, move in opposite directions on parallel tracks with speeds $54$ km/h and $72$ km/h. How long do they take to completely cross each other?
Take train $A$'s direction as positive. Then $v_A = +54$ km/h, $v_B = -72$ km/h. Relative velocity:
$$v_{AB} = v_A - v_B = +54 - (-72) = +126 \text{ km/h}$$
Magnitudes add because directions differ — exactly the structure of NIOS Example 2.3. Convert: $|v_{AB}| = 126 \times \dfrac{5}{18} = 35$ m/s.
"Completely cross" means the tail of one train passes the head of the other, so the total distance to be covered (in the relative frame) is the sum of both lengths: $200 + 200 = 400$ m.
$$t = \dfrac{400 \text{ m}}{35 \text{ m/s}} \approx 11.4 \text{ s}$$
If both trains had been moving in the same direction at $54$ and $72$ km/h, $|v_{rel}|$ would be only $72 - 54 = 18$ km/h $= 5$ m/s, and the crossing would take $400 / 5 = 80$ s — about seven times longer. The same length covered, the same two trains; the geometry of directions sets the timescale.
A police van moving on a highway with a speed of $30$ km/h fires a bullet at a thief's car speeding away in the same direction with a speed of $192$ km/h. If the muzzle speed of the bullet is $150$ m/s, with what speed does the bullet hit the thief's car? (NCERT Exercise 2.14.)
Step 1 — choose positive direction and convert. Take the common direction of motion as positive. Police van speed: $30$ km/h $= 30 \times \dfrac{5}{18} \approx 8.33$ m/s. Thief's car: $192$ km/h $= 192 \times \dfrac{5}{18} \approx 53.33$ m/s. Both positive.
Step 2 — bullet speed in the ground frame. The muzzle speed $150$ m/s is the bullet's speed relative to the van (i.e. in the gun's frame). To get the bullet's velocity relative to the ground, add the van's ground velocity:
$$v_{\text{bullet, ground}} = v_{\text{bullet, van}} + v_{\text{van, ground}} = 150 + 8.33 = 158.33 \text{ m/s}$$
Step 3 — bullet speed relative to the thief's car. This is the speed "relevant for damaging the thief's car" (NCERT's phrasing). Use $v_{\text{bullet, thief}} = v_{\text{bullet, ground}} - v_{\text{thief, ground}}$:
$$v_{\text{bullet, thief}} = 158.33 - 53.33 = 105 \text{ m/s}$$
Answer: the bullet strikes the thief's car at $105$ m/s. The takeaway is that the muzzle speed of $150$ m/s is misleading on its own — what matters for the impact is the bullet's speed in the target's frame, which is smaller because the target is itself moving away.
Position-time graph interpretation
The slope of the position-time graph at any instant equals the velocity at that instant (this is the calculus result of §2.2). When two bodies are plotted on the same set of axes, their relative velocity at any instant is the difference of the two slopes at that instant:
$$v_{BA}(t) = \frac{dx_B}{dt} - \frac{dx_A}{dt}$$Three immediate consequences. First, if the two graphs are parallel straight lines, the two slopes are equal, so $v_{BA} = 0$ — the bodies move with the same velocity, separation constant in time. Second, if the two lines intersect, the bodies are at the same position at the moment of intersection — that is the time-to-meet. The relative velocity at that instant is the difference of the slopes at the intersection point. Third, if one graph is a straight line and the other curves, the relative velocity changes with time because at least one body is accelerating — the relative acceleration equals the difference of the two curvatures (in a calculus sense, the difference of the second derivatives).
Geometrically, you can also plot a single curve of relative position $x_{BA}(t) = x_B(t) - x_A(t)$ against $t$. The slope of this single curve at any instant is $v_{BA}$, exactly as you would expect from the formula. NIOS §2.1.2 derives this relation algebraically: starting from $x_A(t) = x_A(0) + v_A t$ and $x_B(t) = x_B(0) + v_B t$ for uniform motion, the separation $x_{BA}(t) = x_{BA}(0) + v_{BA} t$ — the relative motion is itself uniform, with effective velocity $v_{BA}$.
What this subtopic locked in
- One formula governs everything. $v_{BA} = v_B - v_A$ with signed velocities, in a chosen positive direction.
- Same direction → magnitudes subtract; opposite direction → magnitudes add. Both follow from the signed subtraction automatically.
- $v_{BA} = -v_{AB}$. Subscript order encodes direction; swap subscripts and you flip the sign.
- Time-to-meet shortcut. $t = $ initial separation $/$ $|v_{rel}|$. Crossing two trains uses the sum of their lengths as the separation.
- NCERT police van. Muzzle speed is in the gun's frame; what damages the thief's car is the bullet's speed in the thief's frame — $105$ m/s for the Exercise 2.14 numbers.
- Graph view. $v_{BA} = $ difference of slopes of the two $x$-$t$ graphs. Parallel lines $\Rightarrow$ zero relative velocity. Intersection $\Rightarrow$ instant of meeting.
- Galilean only. Simple subtraction works between inertial frames; accelerating frames need pseudo forces, not in NEET Class 11 kinematics.