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Motion in a Straight Line - Relative Velocity

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Relative velocity is the velocity of an object AA as observed from a frame of reference where another object BB is at rest.

For motion in a straight line, the relative velocity of object AA with respect to object BB is given by the vector difference of their velocities.

If two objects are moving in the same direction, the magnitude of their relative velocity is the difference of their speeds: vrel=vAvBv_{rel} = |v_A - v_B|.

If two objects are moving in opposite directions, the magnitude of their relative velocity is the sum of their speeds: vrel=vA+vBv_{rel} = |v_A + v_B|.

The relative displacement between two objects at any time tt can be expressed as xAB(t)=xAB(0)+vABtx_{AB}(t) = x_{AB}(0) + v_{AB}t, where xAB(0)x_{AB}(0) is the initial separation.

📐Formulae

vAB=vAvBv_{AB} = v_A - v_B

vBA=vBvAv_{BA} = v_B - v_A

vAB=vBAv_{AB} = -v_{BA}

t=Relative DistanceRelative Velocityt = \frac{\text{Relative Distance}}{\text{Relative Velocity}}

💡Examples

Problem 1:

Two trains AA and BB are moving on parallel tracks with velocities 72 km/h72\text{ km/h} and 54 km/h54\text{ km/h} respectively in the same direction. Calculate the relative velocity of AA with respect to BB.

Solution:

vA=72 km/h=72×518=20 m/sv_A = 72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}.\ vB=54 km/h=54×518=15 m/sv_B = 54\text{ km/h} = 54 \times \frac{5}{18} = 15\text{ m/s}.\ vAB=vAvB=2015=5 m/sv_{AB} = v_A - v_B = 20 - 15 = 5\text{ m/s}.

Explanation:

Since both trains are moving in the same direction, we subtract the velocity of the observer (Train BB) from the velocity of the object (Train AA).

Problem 2:

Two cars PP and QQ are 100 m100\text{ m} apart. Car PP moves at 10 m/s10\text{ m/s} and Car QQ moves at 15 m/s15\text{ m/s} towards each other. Find the time when they will meet.

Solution:

Velocity of PP (vPv_P) = 10 m/s10\text{ m/s}.\ Velocity of QQ (vQv_Q) = 15 m/s-15\text{ m/s} (opposite direction).\ Relative velocity vPQ=vPvQ=10(15)=25 m/sv_{PQ} = v_P - v_Q = 10 - (-15) = 25\text{ m/s}.\ Distance s=100 ms = 100\text{ m}.\ Time t=svPQ=10025=4 st = \frac{s}{v_{PQ}} = \frac{100}{25} = 4\text{ s}.

Explanation:

When objects move towards each other, their relative speed increases (sum of magnitudes). The time of meeting is the initial separation divided by this relative speed.

Relative Velocity - Revision Notes & Key Formulas | CBSE Class 11 Physics