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 by differentiating their position equations to find their velocity and acceleration.

### Step-by-Step Solution:

1. **Equation for Particle A:**
\[ x_A = 6t - 3 \]
- **Velocity of A:**
\[ v_A = \frac{dx_A}{dt} = 6 \]
- **Acceleration of A:**
\[ a_A = \frac{dv_A}{dt} = 0 \]
- **Conclusion for A:** Particle A has zero acceleration (constant velocity).

2. **Equation for Particle B:**
\[ x_B = 4t^2 - 2t + 3 \]
- **Velocity of B:**
\[ v_B = \frac{dx_B}{dt} = 8t - 2 \]
- **Acceleration of B:**
\[ a_B = \frac{dv_B}{dt} = 8 \]
- **Conclusion for B:** Particle B has constant acceleration of \(8 \, \text{m/s}^2\).

3. **Equation for Particle C:**
\[ x_C = 3t^3 - 2t^2 + t - 7 \]
- **Velocity of C:**
\[ v_C = \frac{dx_C}{dt} = 9t^2 - 4t + 1 \]
- **Acceleration of C:**
\[ a_C = \frac{dv_C}{dt} = 18t - 4 \]
- **Conclusion for C:** Particle C has time-dependent acceleration (not constant).

4. **Equation for Particle D:**
\[ x_D = 7 \cos(60^\circ) - 3 \sin(30^\circ) \]
- **Calculating constants:**
\[ 7 \cos(60^\circ) = 7 \times \frac{1}{2} = 3.5 \]
\[ 3 \sin(30^\circ) = 3 \times \frac{1}{2} = 1.5 \]
- **Position of D:**
\[ x_D = 3.5 - 1.5 = 2 \]
- **Velocity of D:**
\[ v_D = \frac{dx_D}{dt} = 0 \]
- **Acceleration of D:**
\[ a_D = \frac{dv_D}{dt} = 0 \]
- **Conclusion for D:** Particle D has zero acceleration (at rest).

### Final Conclusion:
The particle moving with constant acceleration is **Particle B** with an acceleration of \(8 \, \text{m/s}^2\).

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