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Physics Question 84 – AP-EAMCET 2026

To a person going towards east in a car with a velocity of 25 kmph, a train appears to move towards north with a velocity of 253 kmph. The actual velocity of the train will be

This problem involves understanding how velocities combine when observed from a moving frame of reference. The observed velocity is the relative velocity, which is the actual velocity of the object minus the velocity of the observer.

Step 1: Identify Given Velocities and the Relative Velocity Equation✦ Active

We are given the velocity of the person (car) with respect to the ground, v→pg, and the velocity of the train as observed by the person, which is the relative velocity of the train with respect to the person, v→tp. We need to find the actual velocity of the train with respect to the ground, v→tg.

The relative velocity equation relating these quantities is:

v→tp=v→tg−v→pg

Rearranging to find the actual velocity of the train:

v→tg=v→tp+v→pg
💡 Teacher's Secret Hint

Remember that 'appears to move' implies relative velocity. Always start by clearly defining your velocity vectors and the frame of reference.

Step 2: Represent Velocities as Vectors○ Expand

Let's define a coordinate system where East is along the positive x-axis (i^) and North is along the positive y-axis (j^). The given velocities are:

Velocity of the person (car) towards East: vpg=25 kmph

v→pg=25i^ kmph

Velocity of the train relative to the person towards North: vtp=253 kmph

v→tp=253j^ kmph
💡 Teacher's Secret Hint

Correctly establishing the vector components based on directions (East, North) is crucial. A simple sketch can help visualize this.

Step 3: Calculate the Actual Velocity of the Train (Vector Sum)○ Expand

Now, substitute the vector forms into the rearranged equation:

v→tg=v→tp+v→pg
v→tg=253j^+25i^
v→tg=25i^+253j^ kmph
💡 Teacher's Secret Hint

Vector addition involves adding corresponding components. Since these vectors are perpendicular, their components remain distinct.

Step 4: Determine the Magnitude of the Actual Velocity of the Train○ Expand

The magnitude of the actual velocity of the train can be found using the Pythagorean theorem, as the x and y components are perpendicular:

|v→tg|=(vtg,x)2+(vtg,y)2
|v→tg|=(25)2+(253)2
|v→tg|=252+252×3
|v→tg|=252(1+3)
|v→tg|=252×4
|v→tg|=25×4
|v→tg|=25×2
|v→tg|=50 kmph
💡 Teacher's Secret Hint

Simplify calculations by factoring out common terms like 252 before taking the square root. This often prevents large number calculations and reduces error.

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