For a particle moving along a straight line, the displacement x depends on time `t` as `x=At^(3)+Bt^(2)+Ct+D`. The ratio of its initial velocity to its initial acceleration depends on:

Options

A. A & C
B. B & C
C. C
D. C and D

Answer

To solve the problem, we need to find the initial velocity and initial acceleration of the particle based on the given displacement equation and then determine the ratio of these two quantities.

### Step-by-Step Solution:

1. **Given Displacement Equation**:
The displacement of the particle is given by:
\[ x = At^3 + Bt^2 + Ct + D \]

2. **Find Initial Velocity**:
The velocity \( v \) is the first derivative of displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(At^3 + Bt^2 + Ct + D) \]
Differentiating term by term:
\[ v = 3At^2 + 2Bt + C \]
To find the initial velocity \( v_0 \), we evaluate \( v \) at \( t = 0 \):
\[ v_0 = 3A(0)^2 + 2B(0) + C = C \]

3. **Find Initial Acceleration**:
The acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(3At^2 + 2Bt + C) \]
Differentiating term by term:
\[ a = 6At + 2B \]
To find the initial acceleration \( a_0 \), we evaluate \( a \) at \( t = 0 \):
\[ a_0 = 6A(0) + 2B = 2B \]

4. **Calculate the Ratio of Initial Velocity to Initial Acceleration**:
Now, we can find the ratio of initial velocity \( v_0 \) to initial acceleration \( a_0 \):
\[ \text{Ratio} = \frac{v_0}{a_0} = \frac{C}{2B} \]

### Final Answer:
The ratio of the initial velocity to the initial acceleration is:
\[ \frac{C}{2B} \]