The potential energy `U` in joule of a particle of mass `1 kg` moving in `x-y` plane obeys the law`U = 3x + 4y`, where `(x,y)` are the co-ordinates of the particle in metre. If the particle is at rest at `(6,4)` at time `t = 0` then :

Options

A. the particle has constant acceleration
B. the particle has zero acceleration
C. the speed of particle when it crosses the y-axis is `10 m//s`
D. co-ordinates of particle at t = 1 s are (4, 5, 2)

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the Force Acting on the Particle
The potential energy \( U \) is given by:
\[ U = 3x + 4y \]
The force \( \mathbf{F} \) acting on the particle can be found using the relation:
\[ \mathbf{F} = -\nabla U \]
where \( \nabla U \) is the gradient of the potential energy. The components of the force can be calculated as:
\[ F_x = -\frac{\partial U}{\partial x} = -3 \]
\[ F_y = -\frac{\partial U}{\partial y} = -4 \]
Thus, the force vector is:
\[ \mathbf{F} = -3 \hat{i} - 4 \hat{j} \]

### Step 2: Calculate the Acceleration of the Particle
Using Newton's second law, \( \mathbf{F} = m \mathbf{a} \), where \( m = 1 \, \text{kg} \):
\[ \mathbf{a} = \frac{\mathbf{F}}{m} = -3 \hat{i} - 4 \hat{j} \]
So, the acceleration of the particle is:
\[ \mathbf{a} = -3 \hat{i} - 4 \hat{j} \]

### Step 3: Determine the Velocity of the Particle After a Certain Time
The particle is at rest at \( (6, 4) \) at \( t = 0 \). Using the kinematic equation:
\[ \mathbf{v} = \mathbf{u} + \mathbf{a} t \]
where \( \mathbf{u} = 0 \) (initial velocity) and \( t = 2 \, \text{s} \):
\[ \mathbf{v} = 0 + (-3 \hat{i} - 4 \hat{j}) \cdot 2 = -6 \hat{i} - 8 \hat{j} \]

### Step 4: Calculate the Speed of the Particle
The speed \( v \) can be calculated using:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s} \]

### Step 5: Find the Position of the Particle at \( t = 1 \, \text{s} \)
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
For the x-coordinate:
\[ x = 6 + 0 \cdot 1 + \frac{1}{2} \cdot (-3) \cdot (1)^2 = 6 - 1.5 = 4.5 \]
For the y-coordinate:
\[ y = 4 + 0 \cdot 1 + \frac{1}{2} \cdot (-4) \cdot (1)^2 = 4 - 2 = 2 \]
Thus, the position of the particle at \( t = 1 \, \text{s} \) is \( (4.5, 2) \).

### Summary of Results
- The force acting on the particle is \( -3 \hat{i} - 4 \hat{j} \).
- The acceleration is \( -3 \hat{i} - 4 \hat{j} \).
- The speed of the particle when it crosses the x-axis is \( 10 \, \text{m/s} \).
- The position of the particle at \( t = 1 \, \text{s} \) is \( (4.5, 2) \).