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Physics · Class 11

Relative velocity

Motion in a Straight Line

Relative position compares the car with the bus rather than with the road origin. Its rate of change is the car velocity minus the bus velocity. If this relative velocity is positive, the car closes an initial gap behind the bus. Changing the observer leaves their relative position unchanged.

Open the interactive experiment →

Subtract velocities to find the closing rate.

பேருந்தைச் சார்ந்த காரின் திசைவேகம், காரின் திசைவேகத்திலிருந்து பேருந்தின் திசைவேகத்தை கழித்த மதிப்பு. தொடக்க இடைவெளி குறையும் வீதத்தை இது காட்டும்.

Key notes, relationships and model limits

Key relationship

vcar/bus = vcar − vbus; xcar/bus = −gap + (vcar − vbus)t

Remember

Relative position compares the car with the bus rather than with the road origin. Its rate of change is the car velocity minus the bus velocity. If this relative velocity is positive, the car closes an initial gap behind the bus. Changing the observer leaves their relative position unchanged.

Common trap

Subtract signed velocities. Opposite-direction speeds add only because their velocity signs differ.

Model limits

Point vehicles; constant signed velocities; car begins at −gap and bus at zero. Vehicles pass without collision. A meeting beyond 10 s is calculated but lies outside the animation.

What you will learn

Calculate relative velocity and meeting time for same and opposite directions; verify observer independence of the gap.

Try the experiment

Switch observers while the cars move. Set equal velocities and see the gap stay fixed; then try opposite directions.

Think of it this way

To catch a moving queue, your progress depends on how much faster you move than the queue.

Worked example

At 5 s the bus is at 30 m and the car is at −20 + 50 = 30 m. In the bus frame both are at zero at that meeting.

Why this matters

Relative motion reduces an overtaking problem to one object approaching a stationary reference.

Before you start

Signed numbers · coordinates · gradients · area under a graph

Check your understanding

A car at +10 m/s is 20 m behind a bus at +6 m/s. When does the car reach the bus?

Read the explanation

Closing rate = 10 − 6 = 4 m/s. Time = 20/4 = 5 s.

Course connection: CBSE XI Chapter 2; NEET and JEE Main Unit 2; JEE Advanced Mechanics. Check the selected syllabus for relative-velocity scope.