Given below are the equations of motion of four particles A, B, C and D `x_(A)=6t-3, x_(B)=4t^(2)-2t+3, x_(C)=3t^(3)-2t^(2)+t-7, x_(D)=7 cos 60^(@)-3 sin 30^(@)`
The particle moving with constant acceleration is

Options

A. A
B. B
C. C
D. D

Answer

To determine which particle is moving with constant acceleration, we need to analyze the equations of motion for each particle and find their respective accelerations. Let's go through each particle step by step.

### Step 1: Analyze Particle A
The equation of motion for particle A is given by:
\[ x_A = 6t - 3 \]

**Velocity Calculation:**
To find the velocity, we differentiate the position function with respect to time \( t \):
\[ v_A = \frac{dx_A}{dt} = \frac{d}{dt}(6t - 3) = 6 \, \text{m/s} \]

**Acceleration Calculation:**
Next, we differentiate the velocity to find the acceleration:
\[ a_A = \frac{dv_A}{dt} = \frac{d}{dt}(6) = 0 \, \text{m/s}^2 \]

### Step 2: Analyze Particle B
The equation of motion for particle B is given by:
\[ x_B = 4t^2 - 2t + 3 \]

**Velocity Calculation:**
Differentiating the position function:
\[ v_B = \frac{dx_B}{dt} = \frac{d}{dt}(4t^2 - 2t + 3) = 8t - 2 \]

**Acceleration Calculation:**
Now, differentiate the velocity:
\[ a_B = \frac{dv_B}{dt} = \frac{d}{dt}(8t - 2) = 8 \, \text{m/s}^2 \]

### Step 3: Analyze Particle C
The equation of motion for particle C is given by:
\[ x_C = 3t^3 - 2t^2 + t - 7 \]

**Velocity Calculation:**
Differentiating the position function:
\[ v_C = \frac{dx_C}{dt} = \frac{d}{dt}(3t^3 - 2t^2 + t - 7) = 9t^2 - 4t + 1 \]

**Acceleration Calculation:**
Now, differentiate the velocity:
\[ a_C = \frac{dv_C}{dt} = \frac{d}{dt}(9t^2 - 4t + 1) = 18t - 4 \]

### Step 4: Analyze Particle D
The equation of motion for particle D is given by:
\[ x_D = 7 \cos(60^\circ) - 3 \sin(30^\circ) \]

Calculating the constants:
\[ x_D = 7 \cdot \frac{1}{2} - 3 \cdot \frac{1}{2} = \frac{7}{2} - \frac{3}{2} = 2 \]

Since \( x_D \) is a constant, we can conclude:
**Velocity Calculation:**
\[ v_D = \frac{dx_D}{dt} = 0 \]

**Acceleration Calculation:**
\[ a_D = \frac{dv_D}{dt} = 0 \]

### Conclusion
From our calculations:
- Particle A has an acceleration of \( 0 \, \text{m/s}^2 \) (no acceleration).
- Particle B has a constant acceleration of \( 8 \, \text{m/s}^2 \).
- Particle C has a time-dependent acceleration \( 18t - 4 \).
- Particle D has an acceleration of \( 0 \, \text{m/s}^2 \) (no acceleration).

Thus, the particle moving with constant acceleration is **Particle B**.