Question

A moving particle has coordinates (5t + 3, 6t, 5)m in frame S at any time 't'. The frame s' is moving with velocity (3\(\hat{i}\)+4\(\hat{j}\))m/s with respect to the frame S. Velocity of particle in frame s' is:

• A. \( 2\hat{i} + 2\hat{j} \)
• B. \( 2\hat{i} - 2\hat{j} \)
• C. \( 3\hat{i} + 4\hat{j} \)
• D. \( 2\hat{i} \)

Hint:

Remember the subscription notation for relative velocity: \( \vec{v}_{A,B} \) means "velocity of A with respect to B". The transformation rule can be written as \( \vec{v}_{A,C} = \vec{v}_{A,B} + \vec{v}_{B,C} \). In this problem, let A = particle, B = frame S', C = frame S. We want \( \vec{v}_{p,S'} \). We know \( \vec{v}_{p,S} \) and \( \vec{v}_{S',S} \). The relation is \( \vec{v}_{p,S} = \vec{v}_{p,S'} + \vec{v}_{S',S} \), which gives the required formula.

Answer 1

The Correct Option is A

Step 1: Conceptual Foundation:
This problem necessitates the application of Galilean velocity transformation, as it concerns relative motion. The provided information includes the particle's position in inertial frame S and the velocity of inertial frame S' relative to S. The objective is to ascertain the particle's velocity as observed from frame S'.
Step 2: Governing Equation:
The Galilean transformation for velocities is defined as:
\[ \vec{v}_{p,S} = \vec{v}_{p,S'} + \vec{v}_{S',S} \]Where:
\(\vec{v}_{p,S}\) is the particle's velocity in frame S.
\(\vec{v}_{p,S'}\) is the particle's velocity in frame S'.
\(\vec{v}_{S',S}\) is the velocity of frame S' relative to frame S.
To determine the velocity in frame S', the formula is rearranged to:
\[ \vec{v}_{p,S'} = \vec{v}_{p,S} - \vec{v}_{S',S} \]Step 3: Calculation Breakdown:
1. Particle Velocity in Frame S (\(\vec{v}_{p,S}\)):
The particle's position vector in frame S is:
\[ \vec{r}_{p,S}(t) = (5t + 3)\hat{i} + (6t)\hat{j} + (5)\hat{k} \]Velocity is obtained by differentiating the position vector with respect to time \( t \):
\[ \vec{v}_{p,S} = \frac{d\vec{r}_{p,S}}{dt} = \frac{d}{dt}[(5t + 3)\hat{i} + (6t)\hat{j} + (5)\hat{k}] \]\[ \vec{v}_{p,S} = 5\hat{i} + 6\hat{j} + 0\hat{k} = (5\hat{i} + 6\hat{j}) \, \text{m/s} \]2. Velocity of Frame S' Relative to S (\(\vec{v}_{S',S}\)):
This value is directly provided:
\[ \vec{v}_{S',S} = (3\hat{i} + 4\hat{j}) \, \text{m/s} \]3. Particle Velocity in Frame S' (\(\vec{v}_{p,S'}\)):
Applying the rearranged transformation formula:
\[ \vec{v}_{p,S'} = \vec{v}_{p,S} - \vec{v}_{S',S} \]\[ \vec{v}_{p,S'} = (5\hat{i} + 6\hat{j}) - (3\hat{i} + 4\hat{j}) \]Performing vector subtraction:
\[ \vec{v}_{p,S'} = (5-3)\hat{i} + (6-4)\hat{j} \]\[ \vec{v}_{p,S'} = (2\hat{i} + 2\hat{j}) \, \text{m/s} \]Step 4: Conclusion:
The calculated velocity of the particle in frame S' is \( 2\hat{i} + 2\hat{j} \) m/s. This result matches option (A).