JEE Physics Relative Motion Notes Set 01

Frames of Reference

Motion is described with respect to an observer, who carries an axis system and a clock. There is no absolute rest: a passenger is at rest for the person beside her in the car but moves backwards past the person on the ground. Whenever we say 'the velocity of A' without naming an observer, the ground is understood. Everything below answers one question: what does one body look like when seen from another, which is itself moving?

Relative Motion on a Line

Subtract the observer's quantity from the object's quantity, each measured from the ground: \( x_{AB}=x_A-x_B \), \( v_{AB}=v_A-v_B \), \( a_{AB}=a_A-a_B \). Here \( v_{AB} \) is the velocity of A as seen by B, and \( v_{BA}=-v_{AB} \). The velocity of anything with respect to itself is zero. Use signs: choose a positive direction and write each velocity with its sign. Two bodies moving towards each other at 10 and 12 m/s have relative speed 22 m/s, while two bodies moving the same way at 20 and 5 m/s have relative speed 15 m/s.
When the relative acceleration is constant, the usual equations apply to the relative quantities: \( v_{rel}=u_{rel}+a_{rel}t \), \( s_{rel}=u_{rel}t+\tfrac12a_{rel}t^2 \), \( v_{rel}^2=u_{rel}^2+2a_{rel}s_{rel} \). Two stones in free fall both have acceleration \( g \), so \( a_{rel}=0 \) and the separation changes uniformly: this is the quickest way to get a meeting time. To avoid a collision just in time, require the relative velocity to fall to zero exactly when the relative displacement equals the initial gap.

Velocity of Approach and Separation

The component of the relative velocity along the line joining two bodies is the velocity of approach if the gap is closing and the velocity of separation if it is opening. In one dimension it is just the magnitude of the relative velocity. In two dimensions, if body 1 moves at angle \( \theta_1 \) to the line joining them and body 2 at \( \theta_2 \) to the same line from the other end, the closing speed is \( v_1\cos\theta_1+v_2\cos\theta_2 \). The gap is smallest at the instant the velocity of approach becomes zero.

Relative Motion in Two Dimensions

The same subtraction is done with vectors: \( \vec v_{AB}=\vec v_A-\vec v_B \). If the two velocities are inclined at angle \( \theta \), \( |\vec v_{AB}|=\sqrt{v_A^2+v_B^2-2v_Av_B\cos\theta} \). For perpendicular velocities of 10 m/s each this gives \( 10\sqrt2 \) m/s. Two bodies in free flight both have acceleration \( \vec g \), so their relative acceleration is zero and each sees the other moving in a straight line at constant velocity. In a lift accelerating upward at \( a \), a body in free flight has acceleration \( g+a \) relative to the lift, so a ball thrown up with speed \( u \) relative to the lift stays up for \( \frac{2u}{g+a} \) and rises \( \frac{u^2}{2(g+a)} \).

Boats and Rivers

Let the swimmer's velocity relative to water be \( \vec v_{mR} \) and the river's velocity relative to ground be \( \vec v_R \). The ground velocity is \( \vec v_m=\vec v_{mR}+\vec v_R \). Along the flow the speed is \( v+u \) downstream and \( v-u \) upstream, so a round trip over distance \( d \) each way takes \( \frac{2dv}{v^2-u^2} \), longer than in still water.
For crossing a river of width \( d \) with the swimmer heading at angle \( \theta \) to the flow: only the component across the river carries the swimmer over, so \( t=\frac{d}{v\sin\theta} \), and the drift downstream is \( x=(u+v\cos\theta)\,t \).
1. Shortest time: head straight across (\( \theta=90^\circ \)), \( t=\frac{d}{v} \), and the drift is \( ud/v \).
2. Shortest path (zero drift): possible only if \( v\ge u \); head upstream so that \( \sin\phi=\frac uv \) with \( \phi \) measured from the perpendicular, giving \( t=\frac{d}{\sqrt{v^2-u^2}} \).
3. If the river is faster than the swimmer, zero drift is impossible; the drift is least when \( \sin\phi=\frac vu \) with \( \phi \) measured from the normal on the upstream side.

Wind and Aeroplanes, Rain and Umbrellas

An aeroplane in wind behaves like a boat in a river: \( \vec v_{a}=\vec v_{aw}+\vec v_{w} \). To keep to a straight track the pilot heads partly into the crosswind. For a trip of length \( \ell \) out and back with a wind of speed \( u \) at angle \( \alpha \) to the track, the time is \( \frac{2\ell\sqrt{v^2-u^2\sin^2\alpha}}{v^2-u^2} \). Around a square of side \( a \) with the wind along one side, the legs take \( \frac{a}{v+u} \), \( \frac{a}{\sqrt{v^2-u^2}} \) (twice) and \( \frac{a}{v-u} \).
Rain is handled the same way: \( \vec v_{rm}=\vec v_r-\vec v_m \). For vertical rain at \( v_r \) and a man walking at \( v_m \), the rain appears to come at angle \( \tan^{-1}\frac{v_m}{v_r} \) to the vertical from the front, and the umbrella is tilted that way. If the man sees vertical rain while moving, the rain's own horizontal speed equals his.

Closest Approach and Collision

Two bodies moving with constant velocities have constant relative velocity, so each moves in a straight line as seen from the other. The closest approach is the perpendicular distance from one body to that line. In symbols, with initial relative position \( \vec r_{12} \) and relative velocity \( \vec v_{12} \): \( d_{min}=\frac{|\vec r_{12}\times\vec v_{12}|}{|\vec v_{12}|} \), reached at \( t=-\frac{\vec r_{12}\cdot\vec v_{12}}{|\vec v_{12}|^2} \). The bodies collide only if the relative velocity points along the line joining them, which means the sideways components of their velocities are equal: \( v_1\sin\theta_1=v_2\sin\theta_2 \); the time is then the gap over the closing speed.
Pursuit puzzles use constant closing speed. If bodies at the corners of a regular polygon of side \( a \) each chase the next at speed \( v \), the closing speed is \( v(1-\cos\frac{2\pi}{n}) \) and they meet after \( \frac{a}{v(1-\cos(2\pi/n))} \): for three bodies \( \frac{2a}{3v} \), for six bodies \( \frac{2a}{v} \).

Worked Examples

Example 1. Two trains move towards each other at 90 km/h and 60 km/h. Relative speed \( 150 \) km/h; if they are 300 km apart they meet after 2 hours.

Example 2. A stone is dropped from 100 m and another is thrown up at 25 m/s from the ground at the same instant (\( g=10 \)). Relative acceleration zero, relative speed 25 m/s, so they meet at \( t=100/25=4 \) s, at height \( 25(4)-5(16)=20 \) m.

Example 3. A boat with speed 5 m/s in still water crosses a 120 m wide river flowing at 3 m/s. Heading straight across: \( t=24 \) s, drift \( 3\times24=72 \) m. To land directly opposite: \( \sin\phi=\frac35 \) and \( t=\frac{120}{\sqrt{25-9}}=30 \) s.

Example 4. A man walks at 4 m/s through rain falling vertically at 3 m/s. The rain relative to him has speed \( \sqrt{16+9}=5 \) m/s at \( \tan^{-1}\frac43 \) from the vertical.

Example 5. Body P at the origin moves at 3 m/s along the x-axis and body Q starts at \( (0,40) \) m moving at 4 m/s in the negative y-direction. \( \vec r_{12}=(0,-40) \), \( \vec v_{12}=(3,4) \), so \( d_{min}=\frac{|0\cdot4-(-40)(3)|}{5}=24 \) m at \( t=\frac{160}{25}=6.4 \) s.