JEE Physics Relative Motion Notes Set 02

Relative Velocity

The velocity of A relative to B is \( \vec v_{AB}=\vec v_A-\vec v_B \), with \( \vec v_{BA}=-\vec v_{AB} \). For motion along a line, subtract the velocities for the same direction and add the magnitudes for opposite directions. Relative acceleration: \( \vec a_{AB}=\vec a_A-\vec a_B \).

Relative Motion with Constant Acceleration

In the frame of B the equations of motion apply with relative velocity and relative acceleration. Two bodies in free fall have zero relative acceleration, so their relative velocity stays constant. Time to meet = initial separation / relative speed (when the relative acceleration is zero).

River-Boat Problems

Let the river speed be \( v_r \), the boat speed in still water \( v_b \) and the river width d. The boat's ground velocity is \( \vec v_b+\vec v_r \).
Shortest time: head perpendicular to the bank, \( t=d/v_b \), with drift \( v_rt \) downstream.
Shortest path (needs \( v_b>v_r \)): head upstream at angle \( \theta \) from the perpendicular with \( \sin\theta=v_r/v_b \), so the crossing speed is \( \sqrt{v_b^2-v_r^2} \) and \( t=\frac{d}{\sqrt{v_b^2-v_r^2}} \).

Rain and Wind Problems

Rain falling vertically with speed \( v_R \) appears to a man walking with speed \( v_M \) to come at angle \( \tan\theta=v_M/v_R \) from the vertical, towards him; the umbrella is held at that angle. A flag flutters along the velocity of the wind relative to the flag pole or vehicle.

Aircraft in Wind

Ground velocity = air velocity + wind velocity. To fly due north in a west-to-east wind of speed w with air speed v, head west of north by \( \sin\theta=w/v \); the ground speed is \( \sqrt{v^2-w^2} \).

Closest Approach

Two particles at separation a with relative velocity having components \( v_1 \) along and \( v_2 \) perpendicular to the line joining them are closest at \( t=\frac{av_1}{v_1^2+v_2^2} \), with minimum separation \( \frac{av_2}{\sqrt{v_1^2+v_2^2}} \).

Worked Example

A boat with speed 5 km/h in still water crosses a river 2 km wide flowing at 3 km/h by the shortest path. Find the time.
Solution: crossing speed = \( \sqrt{25-9}=4 \) km/h, so the time = 2/4 = 0.5 h = 30 minutes.