Relative motion — 1D and 2D problems, rain umbrella problems

Question

How do we solve relative motion problems in 1D and 2D, including the classic rain-umbrella problem?

Solution — Step by Step

The fundamental equation

The velocity of object A relative to object B:

$\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$

This means: "what A's velocity looks like from B's perspective." If you are sitting in B's frame, everything around you appears to move with an additional $-\vec{v}_B$ .

1D relative motion

Two cars on a highway:

Car A moves at 60 km/h east
Car B moves at 40 km/h east

Velocity of A relative to B: $60 - 40 = 20$ km/h east (A appears to move slowly away from B).

Velocity of A relative to B when B moves west at 40 km/h: $60 - (-40) = 100$ km/h east (they approach/separate much faster).

Key rule: Same direction means subtract speeds. Opposite direction means add speeds (for relative approach speed).

2D relative motion — the vector approach

When velocities are not along the same line, we must use vector subtraction.

$\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$

Find the magnitude using: $|\vec{v}_{A/B}| = \sqrt{v_A^2 + v_B^2 - 2v_A v_B \cos\theta}$

where $\theta$ is the angle between $\vec{v}_A$ and $\vec{v}_B$ .

Rain-umbrella problem

Rain falls vertically at speed $v_r$ . You walk horizontally at speed $v_m$ . What angle should you tilt your umbrella?

In your frame (the walking person), rain appears to come at you from an angle. The rain's velocity relative to you:

$\vec{v}_{rain/you} = \vec{v}_{rain} - \vec{v}_{you}$

The rain velocity is $v_r$ (downward) and your velocity is $v_m$ (horizontal). The resultant:

$|\vec{v}_{rain/you}| = \sqrt{v_r^2 + v_m^2}$

Angle with vertical:

$\tan\alpha = \frac{v_m}{v_r}$

You must tilt the umbrella at angle $\alpha$ from vertical, in the direction of your motion.

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Why This Works

Relative motion is about changing your reference frame. When you sit inside a moving train, the platform appears to move backward. Mathematically, subtracting the observer's velocity from everything gives the motion as seen by the observer. This is Galilean relativity — valid at speeds much less than the speed of light.

The rain-umbrella problem is elegant: the rain has no horizontal velocity in the ground frame, but in your frame it acquires a horizontal component equal and opposite to your walking speed.

Alternative Method

For river-crossing problems (a special case of 2D relative motion), use the component method directly. The boat's velocity relative to ground = boat's velocity relative to water + water's velocity. To cross in minimum time, point the boat perpendicular to the river. To cross with zero drift, angle the boat upstream so that the river's current component is exactly cancelled.

Common Mistake

⚠️ Common Mistake

In rain problems, students often calculate $\tan\alpha = v_r / v_m$ (rain speed over man speed). This gives the wrong angle. The correct formula is $\tan\alpha = v_m / v_r$ (man speed over rain speed), where $\alpha$ is measured from the vertical. Remember: if you walk faster, you tilt the umbrella more forward — so $v_m$ must be in the numerator when the angle is from vertical. JEE Main 2022 had this as an option trap.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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