The displacement x of a particle at time t along a straight line is given by `x = alpha - beta t+gamma t^(2)`. The acceleraion of the particle is

Options

A. `-beta`
B. `-beta+2gamma`
C. `2 gamma`
D. `-2gamma`

Explanation

To find the acceleration of the particle given the displacement equation \( x = \alpha - \beta t + \gamma t^2 \), we will follow these steps:

### Step 1: Write down the displacement equation
The displacement of the particle is given by:
\[ x = \alpha - \beta t + \gamma t^2 \]

### Step 2: Differentiate the displacement to find velocity
Velocity \( v \) is the rate of change of displacement with respect to time. Therefore, we differentiate \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(\alpha - \beta t + \gamma t^2) \]
Using the rules of differentiation:
- The derivative of a constant (\(\alpha\)) is 0.
- The derivative of \(-\beta t\) is \(-\beta\).
- The derivative of \(\gamma t^2\) is \(2\gamma t\).

Thus, we have:
\[ v = -\beta + 2\gamma t \]

### Step 3: Differentiate the velocity to find acceleration
Acceleration \( a \) is the rate of change of velocity with respect to time. Therefore, we differentiate \( v \) with respect to \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(-\beta + 2\gamma t) \]
Again, applying the rules of differentiation:
- The derivative of a constant (\(-\beta\)) is 0.
- The derivative of \(2\gamma t\) is \(2\gamma\).

Thus, we have:
\[ a = 2\gamma \]

### Conclusion
The acceleration of the particle is:
\[ \boxed{2\gamma} \]

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