Physics Q84 – AP-EAMCET 2026

Home Physics Motion in a PlaneQuestion #300

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

25 \sqrt{3} kmph

. The actual velocity of the train will be

Clue 1 of 3: Relative Velocity Concept

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.

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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, ${\vec{v}}_{p g}$ , and the velocity of the train as observed by the person, which is the relative velocity of the train with respect to the person, ${\vec{v}}_{t p}$ . We need to find the actual velocity of the train with respect to the ground, ${\vec{v}}_{t g}$ .

The relative velocity equation relating these quantities is:

{\vec{v}}_{t p} = {\vec{v}}_{t g} - {\vec{v}}_{p g}

Rearranging to find the actual velocity of the train:

{\vec{v}}_{t g} = {\vec{v}}_{t p} + {\vec{v}}_{p g}

💡 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 ( $\hat{i}$ ) and North is along the positive y-axis ( $\hat{j}$ ). The given velocities are:

Velocity of the person (car) towards East: $v_{p g} = 25 kmph$

{\vec{v}}_{p g} = 25 \hat{i} kmph

Velocity of the train relative to the person towards North: $v_{t p} = 25 \sqrt{3} kmph$

{\vec{v}}_{t p} = 25 \sqrt{3} \hat{j} 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:

{\vec{v}}_{t g} = {\vec{v}}_{t p} + {\vec{v}}_{p g}

{\vec{v}}_{t g} = 25 \sqrt{3} \hat{j} + 25 \hat{i}

{\vec{v}}_{t g} = 25 \hat{i} + 25 \sqrt{3} \hat{j} 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:

| {\vec{v}}_{t g} | = \sqrt{(v_{t g, x})^{2} + (v_{t g, y})^{2}}

| {\vec{v}}_{t g} | = \sqrt{(25)^{2} + (25 \sqrt{3})^{2}}

| {\vec{v}}_{t g} | = \sqrt{25^{2} + 25^{2} \times 3}

| {\vec{v}}_{t g} | = \sqrt{25^{2} (1 + 3)}

| {\vec{v}}_{t g} | = \sqrt{25^{2} \times 4}

| {\vec{v}}_{t g} | = 25 \times \sqrt{4}

| {\vec{v}}_{t g} | = 25 \times 2

| {\vec{v}}_{t g} | = 50 kmph

💡 Teacher's Secret Hint

Simplify calculations by factoring out common terms like $25^{2}$ before taking the square root. This often prevents large number calculations and reduces error.

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Physics Question 84 – AP-EAMCET 2026

Clue 1 of 3: Relative Velocity Concept

🧠 Full Solution Path