Relative Motion in Two Dimension: Definition | Examples | Practice Questions (original) (raw)

Last Updated : 23 Jul, 2025

Relative motion refers to the speed at which two objects move about each other. It depends on the direction and speed of both objects involved. For example, the speed of a moving vehicle is measured with respect to the ground. The position is also measured with respect to a reference, which is called the origin.

A train moving has a velocity of 100 Km/h with respect to the ground, but if another train is moving at 150 km/h. The velocity of the first train will not be 100 km/h with respect to the person sitting on the second train. It is essential to study the relative motion of the objects.

Reference Frames

In physics, a frame of reference is an abstract coordinate system and a set of physical reference points that uniquely fix the coordinate system help in standardizing the measurements with respect to that frame.

**Example: Consider the figure below. The velocity of the cart with respect to the boy is non-zero, but the velocity of the cart with respect to the person sitting inside the car is zero.

Reference Frames

Frame of Reference and Relative Motion

**Scenario 1:

**Scenario 2:

Relative Motion in Two Dimensions

These concepts can be extended to two-dimensional spaces also. Given in the figure below, consider a particle P and reference frames **S and **S'.

The position of the frame **S' as measured in **S is **r S'S, the position of the particle **P as measured with respect to the frame **S' is given by **r PS' and the position of the particle **P with respect to the frame of reference **S is given by r PS,

Relative Motion in Two Dimensions

**Notice from the figure that,

**r PS = r PS' **+ r S'S

**These vectors give us the formula for relative velocities too, differentiating the above equation,

\frac{d}{dt}(r_{PS}) = \frac{d}{dt}(r_{PS'} + r_{S'S})

⇒\vec{v}_{PS} = \vec{v}_{PS'} + \vec{v}_{S'S}

Intuitively speaking, the velocity of a particle with respect to S is equal to the velocity of S' with respect to S plus the velocity of the particle with respect to S.

**Differentiating this equation again, the equation for the acceleration is given by,

\frac{d}{dt}(\vec{v}_{PS}) = \frac{d}{dt}(\vec{v}_{PS'} + \vec{v}_{S'S})

⇒\vec{a}_{PS} = \vec{a}_{PS'} + \vec{a}_{S'S}

The acceleration of a particle with respect to S is equal to the acceleration of S' with respect to S plus the acceleration of the particle with respect to S.

**Also Read,

Solved Examples of Relative Motion in Two Dimension

**Question 1: A train is moving at a speed of 100 Km/h. The person sitting inside the train starts moving in the direction of the train at a speed of 10 km/h. Find the velocity of the person with respect to the ground. What is the direction of the person's velocity with respect to the ground if they are moving in the same direction as the train?

**Solution:

**Given:

The equation mentioned above states, **v PG = v PT + v TG

plugging the values into the above equation,
⇒ **v PG = 100 + 10
⇒ **v PG = 110

Since, the person is moving in the same direction as the train, the direction of the velocity is the same as the direction in which the train is moving.

**Question 2: A train is moving at a speed of 100 Km/h. The person sitting inside the train starts moving in the opposite direction of the train at a speed of 10 km/h. Find the velocity of the person with respect to the ground.

**Solution:

**Given:

**The equation mentioned above states, **v PG = v PT + v TG

**Plugging the values into the above equation,

⇒ **v PG = 100 - 10
⇒ v PG = 90

**Question 3: A vehicle is moving at a speed of 3i + 4j m/s. The person inside the car thinks the bird is flying at a velocity of 2i + 2j m/s. Find the speed of the bird with respect to the direct.

**Solution:

**Given,

**The equation mentioned above states, v BG = v BV + v VG

**plugging the values into the above equation,

⇒ vPG = 2i + 2j + 3i + 4j
⇒ vPG = 5i + 6j
⇒ |vPG| = √61 m/s

**Question 4: Two particles A and B are moving with velocities 30Km/h and 40Km/h respectively towards an intersection as shown in the figure. Find the velocity of particle A with respect to the velocity of particle B.

Example 4

**Solution:

**Given, v A = 30 Km/h and **v B= 40 Km/h.

Figure shows they are moving in perpendicular directions.

**Using the equation studied above, velocity of A with respect to B: **v AB

Assuming that the earth is the connecting frame of reference here,

\vec{v}_{AB} = \vec{v}_{AE} + \vec{v}_{EB}
⇒ \vec{v}_{AB} = \vec{v}_{AE} - \vec{v}_{BE}
⇒ \vec{v}_{AB} = 30\hat{i} -(-40\hat{j})
⇒\vec{v}_{AB} = 30\hat{i} + 40\hat{j}

Magnitude of this velocity is

\vec{v}_{AB} = 30\hat{i} + 40\hat{j}
⇒ |\vec{v}_{AB}| = \sqrt{30^2 + 40^2}
⇒ |\vec{v}_{AB}| = 50

**Question 5: Two particles A and B are moving with velocities 10Km/h and 20Km/h respectively towards an intersection as shown in the figure. Find the velocity of particle A with respect to the velocity of particle B.

Example 5

**Solution:

**Given: v A = 10 Km/h and **v B = 20 Km/h.

Figure shows they are moving in perpendicular directions.

Using the equation studied above, velocity of A with respect to B: **v AB

Assuming that the earth is the connecting frame of reference here,

\vec{v}_{AB} = \vec{v}_{AE} + \vec{v}_{EB}
⇒ \vec{v}_{AB} = \vec{v}_{AE} - \vec{v}_{BE}
⇒ \vec{v}_{AB} = 10\hat{i} -(-20\hat{j})
⇒\vec{v}_{AB} = 10\hat{i} + 20\hat{j}

Magnitude of this velocity is

\vec{v}_{AB} = 10\hat{i} + 20\hat{j}
⇒ |\vec{v}_{AB}| = \sqrt{10^2 + 20^2}
⇒ |\vec{v}_{AB}| = \sqrt{500}]

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