Relative Velocity in Two Dimensions — River-Boat & Rain (Syllabus Supplement)
1. What Is Relative Velocity in a Plane?
Chapter 2's rule vAB = vA − vB survives intact — only now the subtraction is a genuine vector subtraction (Topic 1), done component by component. The velocity of A as seen by B is vAB = vA − vB; reversing the observer flips the vector: vBA = −vAB; and hopping through an intermediate frame obeys the chain rule vAC = vAB + vBC — the sentence behind every "boat-in-river" and "man-in-rain" problem: vboat,ground = vboat,water + vwater,ground.
Two frame-switching scenarios dominate the exam. The river-boat problem: a boat that can move at vbw relative to the water must cross a river of width d flowing at vw — the question is always aim for what? The rain-man problem: vertically falling rain acquires an apparent horizontal component for a moving observer, dictating the umbrella angle. Both are nothing more than careful vAB = vA − vB bookkeeping with a triangle.
2. Complete Theory: Two Crossing Modes and the Umbrella Angle
Mode 1 — shortest time. The crossing time is governed only by the velocity component perpendicular to the banks: t = d/v⊥. To minimise t, maximise v⊥ — aim the boat straight across (v⊥ = vbw, its maximum possible value). Then tmin = d/vbw, and the current carries the boat a drift = vwtmin downstream. Shortest time always buys a landing point downstream of the intended target.
Mode 2 — zero drift (shortest path). To land exactly opposite, the boat's upstream component must cancel the current: vbw sin φ = vw, where φ is the angle upstream from the perpendicular. This demands vbw > vw; otherwise sin φ would exceed 1 and no heading can save the day. When possible, the effective crossing speed is √(vbw² − vw²) and the time is d/√(vbw² − vw²) — always longer than the shortest time. The comparison "tshortest-path vs tshortest-time" is a stock NEET MCQ pair.
Rain-umbrella. Rain falls vertically at vr; an observer runs horizontally at vm. In the observer's frame the rain's relative velocity is vr,man = vr − vm — a vector with vertical part vr and horizontal part vm directed opposite to the run. The rain therefore appears to come from ahead, and the umbrella must tilt forward from the vertical by θ where tan θ = vm/vr. The relative rain speed is √(vm² + vr²), and when the numbers are 12 and 35 it lands on the Pythagorean triple 37 — NTA's favourite dressing for this classic.
3. Visualising the Zero-Drift Crossing
Figure 3.7 — Aiming upstream so the resultant crosses straight
The cyan arrow is the current; the blue arrow is the boat's heading relative to water (tilted upstream by φ); the green arrow is the actual ground-frame velocity, straight across.
Exam read-out: the 3-4-5 triangle again, rotated into a river. Upstream component 5 sin 37° = 3 cancels the current; the across component is 5 cos 37° = 4. For width 100 m: zero-drift time 25 s vs shortest time 100/5 = 20 s with 60 m drift. Two modes, one triangle.
4. Solved Examples
Example 1 — Both crossing modes on one river
A river 100 m wide flows at 3 m/s. A boat can move at 5 m/s in still water. Find (i) the shortest crossing time and the drift, (ii) the time and heading for landing exactly opposite.
Solution (step by step): (i) Head straight across: tmin = d/vbw = 100/5 = 20 s; drift = vwt = 3 × 20 = 60 m downstream. (ii) Zero drift: sin φ = vw/vbw = 3/5 = 0.6 → φ = 37° upstream from the perpendicular; crossing speed = √(25 − 9) = 4 m/s; t = 100/4 = 25 s. Conclusion worth memorising: buying a perfect landing costs 5 extra seconds on this river.
Example 2 — The 12-35-37 rain problem
Rain falls vertically at 35 m/s. A man walks horizontally at 12 m/s. At what angle from the vertical should he hold his umbrella, and how fast does the rain hit him?
Solution: In the man's frame: vr,man = vr − vm, giving components 35 m/s down and 12 m/s horizontal (opposite his motion — rain comes from ahead). Relative speed = √(12² + 35²) = √(144 + 1225) = √1369 = 37 m/s. Umbrella angle: tan θ = 12/35 = 0.343 → θ ≈ 18.9° from the vertical, tilted forward (into the direction he walks). The umbrella's axis must point opposite the rain's apparent velocity — that is what "shielding" means geometrically.
5. Practice Questions
Q1. A river 60 m wide flows at 4 m/s while a boat manages only 3 m/s in still water. Can the boat land directly opposite?
ANSWER: No. Zero drift requires sin φ = v_w/v_bw = 4/3 > 1 — impossible. The boat will always drift downstream; the best it can do is the straight-across heading (t = 20 s) and then walk back along the bank.
Q2. Rain falls vertically at 8 m/s. A cyclist moves horizontally at 6 m/s. Find the apparent rain speed and the umbrella angle from the vertical.
ANSWER: v_rel = √(6² + 8²) = 10 m/s (the 6-8-10 triple). tan θ = 6/8 = 0.75 → θ ≈ 36.9° from the vertical, tilted forward — the cyclist's 37° shows up again.
Q3 (MCQ). For a boat with v_bw > v_w crossing the same river, which is TRUE? (a) t_shortest-path > t_shortest-time (b) t_shortest-path < t_shortest-time (c) drift is zero in shortest-time mode (d) both times are equal
ANSWER: (a). Shortest-time mode uses full speed across (t = d/v_bw) but drifts; zero-drift mode crosses at the smaller √(v_bw² − v_w²), so it takes longer. (c) is false — drift is v_w t in that mode.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| vAB = vA − vB; vBA = −vAB | Vector subtraction, component-wise |
| vAC = vAB + vBC | Chain rule between frames |
| tmin = d/vbw; drift = vwtmin | Straight-across heading (shortest time) |
| sin φ = vw/vbw; t = d/√(vbw²−vw²) | Zero drift; needs vbw > vw |
| tan θ = vm/vr | Umbrella angle from vertical, tilted forward |
| vrel rain = √(vm² + vr²) | Watch for 3-4-5 triples (12-35-37, 6-8-10) |
The 1-D special case (train overtaking, head-on approach) lives on Chapter 2's relative velocity page — there speeds add or subtract as numbers; here they must be drawn as arrows first.
7. Frequently Asked Questions
How is relative velocity defined in two dimensions?
The velocity of A relative to B (as seen by B) is the vector subtraction v_AB = v_A − v_B, performed component by component. It satisfies the reversal rule v_BA = −v_AB and the chain rule v_AC = v_AB + v_BC, which lets you hop between frames: man-ground = man-water + water-ground. Subtract arrows, never magnitudes — in 2-D the angle between the velocities matters as much as their sizes.
How should a boat aim to cross a river in the shortest time?
For the shortest time, the boat must head straight across, with its full speed v_bw perpendicular to the banks, because the crossing time t = d/v_bw is minimised when the velocity component perpendicular to the flow is maximum. The current then sweeps the boat downstream by drift = v_w t. Shortest time therefore costs a nonzero drift — the boat lands downstream of the point directly opposite.
How can a boat cross a river with zero drift, and when is it impossible?
To land exactly opposite, the boat must aim upstream at an angle φ from the perpendicular such that its upstream component cancels the current: v_bw sin φ = v_w, giving sin φ = v_w/v_bw. The effective crossing speed is v_bw cos φ = √(v_bw² − v_w²), and the time is d/√(v_bw² − v_w²) — longer than the shortest time. Zero drift is impossible whenever v_w ≥ v_bw, because sin φ cannot equal or exceed 1; the boat then necessarily drifts downstream.
Why does an umbrella tilt forward when running in vertically falling rain?
In the runner's frame the rain's relative velocity is v_rain − v_man: the rain acquires a horizontal component opposite to the running direction, so it appears to come from ahead. The umbrella must point into the oncoming rain, i.e., tilt forward from the vertical by θ with tan θ = v_man/v_rain. For rain at 35 m/s and a runner at 12 m/s, the relative rain speed is √(35² + 12²) = 37 m/s and θ = tan⁻¹(12/35) ≈ 18.9°.
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