The Relative Motion Formula gives the velocity or displacement of one object as observed from another moving object, and it is a foundational concept covered in NCERT Class 11 Physics, Chapter 3 (Motion in a Straight Line) and Chapter 4 (Motion in a Plane). This formula is equally critical for JEE Main, JEE Advanced, and NEET, where problems on relative velocity, river-boat motion, and rain-man scenarios appear every year. This article covers every expression, derivation, solved example, and exam tip you need to master relative motion completely.
Key Relative Motion Formulas at a Glance
Quick reference for the most important relative motion expressions used in CBSE and competitive exams.
- Relative velocity of A with respect to B: \( \vec{v}_{AB} = \vec{v}_A – \vec{v}_B \)
- Relative displacement: \( \vec{s}_{AB} = \vec{s}_A – \vec{s}_B \)
- Relative acceleration: \( \vec{a}_{AB} = \vec{a}_A – \vec{a}_B \)
- Magnitude (same direction): \( v_{AB} = v_A – v_B \)
- Magnitude (opposite direction): \( v_{AB} = v_A + v_B \)
- Relative velocity at angle \(\theta\): \( v_{AB} = \sqrt{v_A^2 + v_B^2 – 2v_A v_B \cos\theta} \)
- River-boat crossing time: \( t = \frac{d}{v_{\text{boat}}} \)
What is the Relative Motion Formula?
The Relative Motion Formula describes how the motion of one object appears when observed from another moving object. In everyday life, we always measure motion relative to a fixed reference point, such as the ground. However, when both the observer and the object are moving, we must use relative motion to describe what the observer actually sees.
According to NCERT Class 11 Physics (Chapter 3 and Chapter 4), the velocity of object A as seen by observer B is found by subtracting the velocity of B from the velocity of A. This subtraction is vectorial, meaning both magnitude and direction matter.
The concept is rooted in Galilean relativity. Newton’s second law holds true in any inertial frame of reference. An inertial frame is one that moves at a constant velocity. When two observers move at constant velocities relative to each other, both observe the same physical laws. This principle is essential for solving collision problems, river-boat problems, and rain-man problems in CBSE and competitive exams.
The relative motion formula applies in one dimension (straight-line motion) and two dimensions (plane motion). Understanding both cases is necessary for Class 11, Class 12, JEE, and NEET preparation.
Relative Motion Formula — Expression and Variables
The general vector form of the relative velocity formula is:
\[ \vec{v}_{AB} = \vec{v}_A – \vec{v}_B \]
This reads as: “the velocity of A relative to B equals the velocity of A minus the velocity of B.” Similarly, the relative displacement and relative acceleration formulas follow the same pattern:
\[ \vec{s}_{AB} = \vec{s}_A – \vec{s}_B \]
\[ \vec{a}_{AB} = \vec{a}_A – \vec{a}_B \]
When two objects move at an angle \(\theta\) to each other, the magnitude of relative velocity is:
\[ |\vec{v}_{AB}| = \sqrt{v_A^2 + v_B^2 – 2v_A v_B \cos\theta} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \vec{v}_{AB} \) | Velocity of A relative to B | m/s |
| \( \vec{v}_A \) | Velocity of object A (in ground frame) | m/s |
| \( \vec{v}_B \) | Velocity of object B (in ground frame) | m/s |
| \( \vec{s}_{AB} \) | Displacement of A relative to B | m |
| \( \vec{a}_{AB} \) | Acceleration of A relative to B | m/s² |
| \( \theta \) | Angle between velocity vectors of A and B | degree or radian |
Derivation of the Relative Motion Formula
Consider two objects A and B observed from a fixed ground frame (O). Let the position vectors of A and B from O be \( \vec{r}_A \) and \( \vec{r}_B \) respectively.
Step 1: The position of A as seen from B is: \( \vec{r}_{AB} = \vec{r}_A – \vec{r}_B \)
Step 2: Differentiate both sides with respect to time:
\[ \frac{d\vec{r}_{AB}}{dt} = \frac{d\vec{r}_A}{dt} – \frac{d\vec{r}_B}{dt} \]
Step 3: By definition, \( \frac{d\vec{r}}{dt} = \vec{v} \). Therefore:
\[ \vec{v}_{AB} = \vec{v}_A – \vec{v}_B \]
Step 4: Differentiating once more gives the relative acceleration: \( \vec{a}_{AB} = \vec{a}_A – \vec{a}_B \). This completes the derivation.
Complete Relative Motion Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Relative Velocity (1D, same direction) | \( v_{AB} = v_A – v_B \) | v_A, v_B = speeds of A and B | m/s | Class 11, Ch 3 |
| Relative Velocity (1D, opposite direction) | \( v_{AB} = v_A + v_B \) | v_A, v_B = speeds in opposite directions | m/s | Class 11, Ch 3 |
| Relative Velocity (2D, vector form) | \( \vec{v}_{AB} = \vec{v}_A – \vec{v}_B \) | Vector velocities of A and B | m/s | Class 11, Ch 4 |
| Magnitude at angle \(\theta\) | \( |\vec{v}_{AB}| = \sqrt{v_A^2 + v_B^2 – 2v_A v_B \cos\theta} \) | \(\theta\) = angle between velocity vectors | m/s | Class 11, Ch 4 |
| Relative Displacement | \( \vec{s}_{AB} = \vec{s}_A – \vec{s}_B \) | Position vectors from ground frame | m | Class 11, Ch 3 |
| Relative Acceleration | \( \vec{a}_{AB} = \vec{a}_A – \vec{a}_B \) | Accelerations of A and B in ground frame | m/s² | Class 11, Ch 3 |
| River-Boat: Minimum Drift Angle | \( \sin\alpha = \frac{v_{\text{river}}}{v_{\text{boat}}} \) | \(\alpha\) = angle upstream; drift = 0 | degree | Class 11, Ch 4 |
| River-Boat: Shortest Time Crossing | \( t = \frac{d}{v_{\text{boat}}} \) | d = river width, v_boat = speed perpendicular to river | s | Class 11, Ch 4 |
| Rain-Man: Apparent Velocity of Rain | \( \vec{v}_{\text{rain,man}} = \vec{v}_{\text{rain}} – \vec{v}_{\text{man}} \) | Vector velocities of rain and man | m/s | Class 11, Ch 4 |
| Time to Catch (1D, same direction) | \( t = \frac{s_0}{v_A – v_B} \) | s_0 = initial separation, v_A > v_B | s | Class 11, Ch 3 |
Relative Motion Formula — Solved Examples
Example 1 (Class 9-10 Level): Two Trains Moving in the Same Direction
Problem: Train A moves at 80 km/h and Train B moves at 50 km/h in the same direction on parallel tracks. Find the velocity of Train A relative to Train B.
Given: \( v_A = 80 \) km/h, \( v_B = 50 \) km/h, same direction.
Step 1: Use the relative velocity formula for the same direction: \( v_{AB} = v_A – v_B \)
Step 2: Substitute values: \( v_{AB} = 80 – 50 = 30 \) km/h
Step 3: The direction is the same as the direction of motion of Train A.
Answer
The velocity of Train A relative to Train B is 30 km/h in the direction of motion of Train A. To a passenger in Train B, Train A appears to move forward at 30 km/h.
Example 2 (Class 11-12 Level): Rain-Man Problem
Problem: Rain falls vertically at 20 m/s. A man walks horizontally at 5 m/s towards the east. Find the magnitude and direction of the apparent velocity of rain as experienced by the man. At what angle should he hold his umbrella?
Given: \( \vec{v}_{\text{rain}} = 20 \) m/s (downward, i.e., \(-\hat{j}\) direction), \( \vec{v}_{\text{man}} = 5 \) m/s (east, i.e., \(+\hat{i}\) direction).
Step 1: Apply the relative velocity formula: \( \vec{v}_{\text{rain, man}} = \vec{v}_{\text{rain}} – \vec{v}_{\text{man}} \)
Step 2: Substitute: \( \vec{v}_{\text{rain, man}} = (0\hat{i} – 20\hat{j}) – (5\hat{i} + 0\hat{j}) = -5\hat{i} – 20\hat{j} \)
Step 3: Find the magnitude:
\[ |\vec{v}_{\text{rain, man}}| = \sqrt{(-5)^2 + (-20)^2} = \sqrt{25 + 400} = \sqrt{425} \approx 20.6 \text{ m/s} \]
Step 4: Find the angle \(\phi\) with the vertical:
\[ \tan\phi = \frac{5}{20} = 0.25 \implies \phi = \tan^{-1}(0.25) \approx 14.04° \]
Step 5: The man must tilt his umbrella at approximately 14° towards the east (the direction he is walking).
Answer
The apparent velocity of rain is approximately 20.6 m/s. The umbrella must be tilted at 14.04° towards the east from the vertical.
Example 3 (JEE/NEET Level): River-Boat Problem with Minimum Drift
Problem: A river flows at 4 m/s towards the east. A boat can travel at 5 m/s in still water. The river is 100 m wide. (a) At what angle upstream should the boat be directed to cross with zero drift? (b) How long does the crossing take in this case?
Given: \( v_{\text{river}} = 4 \) m/s (east), \( v_{\text{boat}} = 5 \) m/s (in still water), \( d = 100 \) m.
Step 1: For zero drift, the boat must aim upstream. The boat’s velocity component along the river must cancel the river’s velocity. Let \(\alpha\) be the angle the boat makes with the direction perpendicular to the river (i.e., upstream from north).
Step 2: Apply the zero-drift condition:
\[ v_{\text{boat}} \sin\alpha = v_{\text{river}} \implies \sin\alpha = \frac{4}{5} = 0.8 \implies \alpha = 53.13° \]
Step 3: The effective velocity perpendicular to the river (used for crossing) is:
\[ v_{\perp} = v_{\text{boat}} \cos\alpha = 5 \times \cos(53.13°) = 5 \times 0.6 = 3 \text{ m/s} \]
Step 4: Time to cross:
\[ t = \frac{d}{v_{\perp}} = \frac{100}{3} \approx 33.3 \text{ s} \]
Answer
(a) The boat must be directed at 53.13° upstream from the perpendicular to the river bank. (b) The crossing time is approximately 33.3 seconds.
CBSE Exam Tips 2025-26
- Always define a reference frame first. Before applying any formula, clearly state which object is the observer and which is the observed. This prevents sign errors in the subtraction \( \vec{v}_{AB} = \vec{v}_A – \vec{v}_B \).
- Draw vector diagrams. For 2D problems (rain-man, river-boat), always sketch the velocity triangle. Visual representation helps you identify the correct angle and apply the cosine rule accurately.
- Memorise the two 1D cases. Same direction: subtract velocities. Opposite direction: add velocities. These two rules cover the majority of CBSE short-answer questions.
- Practice the river-boat problem thoroughly. We recommend solving at least 10 river-boat variations, including minimum-time and minimum-drift cases. These appear consistently in CBSE Class 11 annual exams.
- Check units before substituting. Convert km/h to m/s by multiplying by \( \frac{5}{18} \). Mixed units are a common source of errors in board exams.
- Use the cosine rule for the general angle case. When the angle between velocities is given, apply \( |\vec{v}_{AB}| = \sqrt{v_A^2 + v_B^2 – 2v_A v_B \cos\theta} \) directly. This formula is given in NCERT and is fully acceptable in CBSE 2025-26 exams.
Common Mistakes to Avoid
- Reversing the subtraction order: Many students write \( \vec{v}_{AB} = \vec{v}_B – \vec{v}_A \) instead of \( \vec{v}_A – \vec{v}_B \). Remember: the subscript on the left (A) is the object being observed; the subscript on the right (B) is the observer. Subtract the observer’s velocity.
- Treating vectors as scalars in 2D problems: In rain-man and river-boat problems, velocities are vectors. You cannot simply add or subtract their magnitudes. Always resolve into components or use the cosine rule.
- Confusing minimum-time and minimum-drift conditions: For minimum crossing time, the boat must be directed perpendicular to the bank. For zero drift, the boat must aim upstream at angle \( \sin^{-1}(v_{\text{river}}/v_{\text{boat}}) \). These are two different scenarios with different angles.
- Ignoring the direction of relative velocity: Calculating only the magnitude is not enough. CBSE and JEE questions often ask for the direction as well. Always find the angle using \( \tan^{-1} \) after resolving components.
- Applying relative motion to non-inertial frames: The formula \( \vec{v}_{AB} = \vec{v}_A – \vec{v}_B \) is valid only when both frames are inertial (moving at constant velocity). If observer B is accelerating, pseudo forces must be considered separately.
JEE/NEET Application of the Relative Motion Formula
In our experience, JEE aspirants encounter the relative motion formula in at least 2-3 questions per paper, spanning mechanics, kinematics, and even optics (relative motion of mirrors and objects). NEET questions are typically more straightforward but still test the rain-man and river-boat scenarios regularly.
Pattern 1: Relative Velocity in Collisions
JEE Advanced problems often involve two objects approaching each other. The relative velocity of approach is \( v_{\text{approach}} = v_A + v_B \) when they move in opposite directions. After a perfectly elastic collision between equal masses, the velocities exchange. This is most elegantly solved in the centre-of-mass frame, where the total momentum is zero and the relative velocity simply reverses direction.
Pattern 2: River-Boat with Minimum Drift
This is a classic JEE Main topic. The key insight is that minimum drift does NOT mean zero drift (unless \( v_{\text{boat}} \geq v_{\text{river}} \)). When \( v_{\text{boat}} < v_{\text{river}} \), the minimum drift angle is found by differentiating the drift expression with respect to the boat’s heading angle and setting it to zero. This yields a more complex result than the simple zero-drift formula.
Pattern 3: Projectile Seen from Another Projectile
A favourite JEE Advanced concept: if two projectiles are launched simultaneously, their relative acceleration is \( \vec{a}_{AB} = \vec{a}_A – \vec{a}_B = \vec{g} – \vec{g} = 0 \). This means each projectile sees the other moving in a straight line at constant relative velocity. This elegant result appears in multiple JEE papers and is directly derived from the relative acceleration formula.
Our experts suggest practising all three patterns with timed mock questions. Recognising the pattern quickly saves crucial minutes in JEE and NEET.
FAQs on Relative Motion Formula
For deeper understanding, explore our related articles on the Physics Formulas hub, the Critical Velocity Formula, and the Superposition Formula. You can also refer to the official NCERT website for Class 11 Physics textbook chapters on kinematics and motion in a plane.