Relative Velocity

Concept of Relative Velocity

Motion is always relative. The velocity of an object depends on the frame of reference from which it is observed. Relative velocity is the velocity of an object with respect to another object (which can be stationary or moving).

Formulation in One Dimension

If two objects A and B are moving along a straight line with velocities $v_A$ and $v_B$ (with respect to a common reference frame, like the ground), then:

• The velocity of object A with respect to object B is: $$v_{AB} = v_A - v_B$$
• The velocity of object B with respect to object A is: $$v_{BA} = v_B - v_A$$

Note that $$v_{AB} = -v_{BA}$$.

Scenarios

1. Objects moving in the same direction: If both $v_A$ and $v_B$ are positive, $v_{AB}$ will be smaller than $v_A$. If $v_A > v_B$, $v_{AB}$ is positive, meaning A is moving away from B. If $v_A < v_B$, $v_{AB}$ is negative, meaning A is approaching B.
2. Objects moving in opposite directions: If $v_A$ is positive and $v_B$ is negative, then $v_{AB} = v_A - (-v_B) = v_A + v_B$. The magnitude of their relative velocity is the sum of their individual speeds. This is the velocity of approach.

Example:

Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of $72\ km/h$ in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by $1\ m/s^2$. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Solution:

1. Convert initial speeds to m/s: $v_A = v_B = 72\ km/h = 72 \times \frac{5}{18} = 20\ m/s$.

2. Use relative motion concepts: Let's analyze the motion of train B with respect to train A. Initial relative velocity of B w.r.t A: $u_{BA} = v_B - v_A = 20 - 20 = 0$. Relative acceleration of B w.r.t A: $a_{BA} = a_B - a_A = 1 - 0 = 1\ m/s^2$.

3. Calculate relative displacement: For the guard of B to just pass the driver of A, train B must cover its own length plus the length of train A, in addition to the initial distance between them. The relative distance covered is the sum of their lengths, which is $400 + 400 = 800\ m$. Let the initial distance be x. Total relative displacement required to overtake is $s_{rel} = x + 800$. Wait, this is not quite right. For the guard of B to pass the driver of A, the front of B must travel the initial distance (x) plus the length of A (400m). No, the guard is at the back of B. The driver is at the front of A. So B must cover its own length (400m), the initial distance (x), and the length of A (400m). No, that's not right. The total distance for the guard of B to be at the same position as the driver of A is the initial distance x plus the length of train A. Let's reconsider. The front of B must move a distance equal to the initial gap plus the length of A. The guard is 400m behind the front of B. So the guard moves a distance equal to the initial gap. Let's simplify. Relative distance covered by B to overtake A is the sum of their lengths. $s_{rel} = 400 + 400 = 800m$. Let the initial distance between the front of B and back of A be x. To overtake, the front of B must cover distance x+400. To pass completely, it must cover x+400+400. The question is 'guard of B just brushes past the driver of A'. This means the back of B has reached the front of A. The total relative distance covered is the sum of their lengths: $s_{rel} = 400m + 400m = 800m$. Using the kinematic equation for relative motion: $s_{rel} = u_{BA}t + \frac{1}{2}a_{BA}t^2$. $800 = (0)(50) + \frac{1}{2}(1)(50)^2 = \frac{1}{2}(2500) = 1250\ m$. This is the total relative distance. The original distance between them was this total distance minus the length of the trains. No, this $s_{rel}$ is the original distance. Let's re-read. 'guard of B just brushes past the driver of A'. This means the back of B has moved to the position of the front of A. The relative displacement is the original distance between them. So, the original distance was 1250 m.