Relative Motion — When to Use Which Frame

Relative Motion — When to Use Which Frame

• 8 min read
Tags Relative Motion Intuition

Opening — Why Relative Motion Trips Students Up

Relative motion is one of those topics where the mathematics is trivial — vector subtraction — but the choice of frame decides whether a problem takes 30 seconds or 30 minutes. We’ve seen students brute-force a river-crossing problem with calculus when a frame-shift would have given the answer in two lines.

The whole game is this: pick a frame in which one body is stationary, then describe everything else relative to that frame. A boat crossing a river? Sit on the river. Two cars approaching? Sit on one car. A coin tossed in a moving train? Sit on the train. Once you sit on the right body, the problem turns into something familiar — projectile from a stationary platform, or one-dimensional kinematics — that you’ve already solved a hundred times.

Let’s build the intuition properly, starting from the definitions.

Key Terms & Definitions

Frame of reference: a coordinate system attached to some observer. Every measurement of position and velocity depends on the frame.

Velocity of A relative to B, written v⃗AB\vec{v}_{AB} or v⃗A/B\vec{v}_{A/B}:

v⃗AB=v⃗A−v⃗B\vec{v}_{AB} = \vec{v}_A - \vec{v}_B

This reads as: “A’s velocity, as seen by an observer moving with B.”

Position of A relative to B:

r⃗AB=r⃗A−r⃗B\vec{r}_{AB} = \vec{r}_A - \vec{r}_B

Acceleration of A relative to B: a⃗AB=a⃗A−a⃗B\vec{a}_{AB} = \vec{a}_A - \vec{a}_B (only when both frames are inertial, or when we’re careful about pseudo-forces).

v⃗AB=v⃗A−v⃗B\vec{v}_{AB} = \vec{v}_A - \vec{v}_B

v⃗A=v⃗AB+v⃗B(rearrangement)\vec{v}_A = \vec{v}_{AB} + \vec{v}_B \quad \text{(rearrangement)}

Methods — Which Frame to Use

Method 1: River-Boat Problems

A boat with velocity v⃗br\vec{v}_{br} relative to river, river flowing with v⃗r\vec{v}_r relative to ground. Boat’s velocity relative to ground:

v⃗b=v⃗br+v⃗r\vec{v}_b = \vec{v}_{br} + \vec{v}_r

Decision rule: if the question asks about the boat’s path on the water (e.g., “how long to reach the other bank?”), sit on the river. The river-frame velocity of the boat is just v⃗br\vec{v}_{br}, and the bank moves toward the boat at −v⃗r-\vec{v}_r. Crossing time = (river width) / (component of v⃗br\vec{v}_{br} perpendicular to the bank).

If the question asks “where does the boat land on the opposite bank?”, sit on the ground. The boat drifts downstream by v⃗r⋅t\vec{v}_r \cdot t, where tt is the crossing time.

Method 2: Two-Body Approach in Pulleys / Collisions

Two objects of masses m1m_1 and m2m_2 connected by a string over a pulley, or approaching each other on a track — sit on one object. Define v⃗12=v⃗1−v⃗2\vec{v}_{12} = \vec{v}_1 - \vec{v}_2 and a⃗12=a⃗1−a⃗2\vec{a}_{12} = \vec{a}_1 - \vec{a}_2. Treat one object as stationary and let the other move with relative velocity. Half the equations vanish.

Method 3: Rain-Man Problems

Rain falls vertically with speed vrv_r. A person walks horizontally with speed vpv_p. Rain’s velocity relative to person:

v⃗rp=v⃗r−v⃗p\vec{v}_{rp} = \vec{v}_r - \vec{v}_p

The angle at which the person should tilt the umbrella from the vertical:

tan⁡θ=vpvr\tan\theta = \frac{v_p}{v_r}

Direction: against the direction of motion.

Method 4: Projectile from Moving Platform

If a projectile is launched from a moving train, sit on the train (which moves uniformly, so it’s inertial). In the train frame, the projectile is a normal vertical-up-and-down. Then add the train’s velocity to convert back to ground frame.

Solved Examples

Example 1 (Easy — CBSE)

A river is 200 m wide. Boat speed in still water = 4 m/s, river current = 3 m/s. Find the shortest time to cross and the drift downstream.

For shortest crossing time, point the boat perpendicular to the current. Time:

t=2004=50 st = \frac{200}{4} = 50 \text{ s}

Drift downstream:

d=vr⋅t=3⋅50=150 md = v_r \cdot t = 3 \cdot 50 = 150 \text{ m}

Example 2 (Medium — JEE Main 2023)

Two cars A and B move on perpendicular roads with speeds 30 m/s (north) and 40 m/s (east). Find the velocity of A relative to B and its magnitude.

v⃗AB=v⃗A−v⃗B=(0i^+30j^)−(40i^+0j^)=−40i^+30j^\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (0\hat{i} + 30\hat{j}) - (40\hat{i} + 0\hat{j}) = -40\hat{i} + 30\hat{j}

Magnitude:

∣v⃗AB∣=402+302=50 m/s|\vec{v}_{AB}| = \sqrt{40^2 + 30^2} = 50 \text{ m/s}

Direction: north of west.

Example 3 (Hard — JEE Advanced)

Rain falls vertically at 10 m/s. A man walks east at 5 m/s. Rain appears to fall at what angle to the vertical, in his frame? At what angle should he tilt his umbrella?

Rain velocity relative to man: v⃗rm=(0,−10)−(5,0)=(−5,−10)\vec{v}_{rm} = (0, -10) - (5, 0) = (-5, -10).

Angle from vertical: tan⁡θ=5/10=0.5\tan\theta = 5/10 = 0.5, so θ=tan⁡−1(0.5)≈26.57°\theta = \tan^{-1}(0.5) \approx 26.57° west of vertical.

Umbrella tilt: 26.57° away from his direction of motion (i.e., toward the west, since rain comes from the west in his frame).

Exam-Specific Tips

JEE Main: 1-2 questions on relative motion every alternate year, usually river-boat or two-projectile timing. NEET: rare, but appears in conceptual passages on collisions and Doppler. CBSE Class 11: derive boat-river formulas; the board exam pattern almost always asks “shortest distance” or “shortest time” framing.

Common Mistakes to Avoid

Mistake 1: Mixing reference frames mid-problem. Once you’ve decided to sit on body B, every velocity you write must be in B’s frame. Don’t switch halfway.

Mistake 2: Treating non-inertial frames as inertial. If you sit on an accelerating car, you must add a pseudo-force −ma⃗car-m\vec{a}_{\text{car}} to every body. JEE Advanced loves to test this.

Mistake 3: Forgetting that relative position is also a vector. The minimum distance between two moving particles is found from ∣r⃗12∣|\vec{r}_{12}|, not from one Cartesian coordinate.

Mistake 4: Subtraction order. v⃗AB=v⃗A−v⃗B\vec{v}_{AB} = \vec{v}_A - \vec{v}_B, not v⃗B−v⃗A\vec{v}_B - \vec{v}_A. The first letter in the subscript is the “moving” body, the second is the observer.

Mistake 5: Using the wrong angle for shortest path vs shortest time. Shortest time → perpendicular to bank. Shortest path → angle that makes net velocity perpendicular to bank.

Practice Questions

Q1. Two trains, 100 m and 150 m long, moving in opposite directions at 36 km/h and 54 km/h. How long to cross each other completely?

Relative speed = 36+54=9036 + 54 = 90 km/h =25= 25 m/s. Total distance = 100+150=250100 + 150 = 250 m. Time =250/25=10= 250/25 = 10 s.

Q2. A river 500 m wide flows at 5 m/s. A boat can move at 13 m/s in still water. Find the angle at which the boat must be steered to reach a point directly opposite.

For the boat to land directly opposite, the upstream component of v⃗br\vec{v}_{br} must cancel the river current. So sin⁡θ=5/13\sin\theta = 5/13, θ=sin⁡−1(5/13)≈22.6°\theta = \sin^{-1}(5/13) \approx 22.6° upstream from perpendicular.

Q3. A man walks at 4 m/s east; rain appears to fall at 30°30° east of vertical. If the man’s speed doubles, at what angle does the rain appear?

Original: tan⁡30°=vm/vr=4/vr\tan 30° = v_m/v_r = 4/v_r, so vr=4/tan⁡30°=43v_r = 4/\tan 30° = 4\sqrt{3} m/s. With doubled vm=8v_m = 8 m/s: tan⁡θ=8/(43)=2/3\tan\theta = 8/(4\sqrt{3}) = 2/\sqrt{3}, so θ≈49.1°\theta \approx 49.1° east of vertical.

Q4. Two cars approach an intersection at speeds 20 m/s. They are 100 m and 75 m from the intersection. Find the closest distance between them.

Use relative motion: in the frame of one car, the other has relative velocity 20220\sqrt{2} m/s along a direction 45°45° from the line joining them. Closest distance = perpendicular component of initial position vector relative to relative velocity. Working it out gives approximately 17.717.7 m.

Q5. A boy throws a ball straight up at 10 m/s while running at 5 m/s on level ground. Where does the ball land relative to him?

In the boy’s frame (uniform motion, inertial), the ball goes straight up and comes straight down. So the ball lands in his hand. The ground observer sees a parabolic trajectory but the boy keeps pace.

FAQs

Q: Is relative motion only for vectors? Yes — speed (scalar) doesn’t have a direction, so “relative speed” is ambiguous unless directions are specified. Always work with velocity vectors first, then compute magnitudes.

Q: When can I use the relative motion shortcut for collisions? Whenever you’re tracking how far/long it takes for two objects to meet or separate. Define one as the reference, treat the other’s velocity as relative.

Q: How does this relate to special relativity? Galilean (classical) relative motion adds velocities directly. Special relativity uses the relativistic velocity addition formula at speeds close to cc. For JEE/NEET, only Galilean.

Q: Why does the umbrella tilt forward, not backward? Because rain in your frame seems to come from in front of you. To intercept it, you tilt the umbrella into where the rain appears to be coming from.

Q: Can I solve every relative-motion problem without using the relative-motion concept? Yes, but it’s harder. Ground-frame solutions are valid but often require simultaneous equations. Frame-shifting collapses them into one-dimensional problems.

Q: What about angular relative motion? ω⃗AB=ω⃗A−ω⃗B\vec{\omega}_{AB} = \vec{\omega}_A - \vec{\omega}_B — same idea. Relevant for rotating frames and rolling motion.

Q: How do I know the question wants relative or absolute motion? Look at the verbs: “approach”, “separate”, “catch up”, “fall behind” — these signal relative motion. “Reach”, “land at”, “displacement from origin” — usually ground frame.