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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

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

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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.

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

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

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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=vA−vBv_{AB} = v_A - v_B

vBA=vB−vAv_{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=vA−vB=20−15=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=vP−vQ=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.