For a particle moving along a straight line, the displacement x depends on time t as `x = alpha t^(3) +beta t^(2) +gamma t +delta`. The ratio of its initial acceleration to its initial velocity depends

Options

A. only on `alpha` and `gamma`
B. only on `beta` and `gamma`
C. only on `alpha` and `beta`
D. only on `alpha`

Explanation

To solve the problem, we need to find the ratio of the initial acceleration to the initial velocity for the given displacement function \( x(t) = \alpha t^3 + \beta t^2 + \gamma t + \delta \).

### Step 1: Find the velocity function
The velocity \( v(t) \) is the first derivative of the displacement \( x(t) \) with respect to time \( t \).

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(\alpha t^3 + \beta t^2 + \gamma t + \delta) \]

Using the power rule of differentiation, we get:

\[ v(t) = 3\alpha t^2 + 2\beta t + \gamma \]

### Step 2: Calculate initial velocity
To find the initial velocity, we evaluate \( v(t) \) at \( t = 0 \):

\[ v(0) = 3\alpha(0)^2 + 2\beta(0) + \gamma = \gamma \]

Thus, the initial velocity \( v_0 = \gamma \).

### Step 3: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity \( v(t) \) with respect to time \( t \):

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3\alpha t^2 + 2\beta t + \gamma) \]

Again, using the power rule of differentiation, we get:

\[ a(t) = 6\alpha t + 2\beta \]

### Step 4: Calculate initial acceleration
To find the initial acceleration, we evaluate \( a(t) \) at \( t = 0 \):

\[ a(0) = 6\alpha(0) + 2\beta = 2\beta \]

Thus, the initial acceleration \( a_0 = 2\beta \).

### Step 5: Find the ratio of initial acceleration to initial velocity
Now, we can find the ratio of the initial acceleration to the initial velocity:

\[ \text{Ratio} = \frac{a_0}{v_0} = \frac{2\beta}{\gamma} \]

### Conclusion
The ratio of the initial acceleration to the initial velocity is:

\[ \frac{2\beta}{\gamma} \]

This ratio depends only on the coefficients \( \beta \) and \( \gamma \).