The displacement 'x' of a particle moving along a straight line at time t is given by `x=a_(0)+a_(1)t+a_(2)t^(2)`. The acceleration of the particle is :-

Options

A. `a_(1)`
B. `a_(2)`
C. `2a_(2)`
D. `3a_(2)`

Answer

To find the acceleration of the particle given the displacement equation \( x = a_0 + a_1 t + a_2 t^2 \), we will follow these steps:

### Step 1: Differentiate the displacement equation to find velocity.
The displacement \( x \) is given by:
\[ x = a_0 + a_1 t + a_2 t^2 \]
To find the velocity \( v \), we differentiate \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(a_0 + a_1 t + a_2 t^2) \]
Since \( a_0 \) is a constant, its derivative is 0. The derivatives of the other terms are:
\[ \frac{d}{dt}(a_1 t) = a_1 \quad \text{and} \quad \frac{d}{dt}(a_2 t^2) = 2a_2 t \]
Thus, the velocity \( v \) is:
\[ v = a_1 + 2a_2 t \]

### Step 2: Differentiate the velocity equation to find acceleration.
Now, we differentiate the velocity \( v \) with respect to time \( t \) to find the acceleration \( a \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(a_1 + 2a_2 t) \]
Again, since \( a_1 \) is a constant, its derivative is 0. The derivative of \( 2a_2 t \) is:
\[ \frac{d}{dt}(2a_2 t) = 2a_2 \]
Thus, the acceleration \( a \) is:
\[ a = 2a_2 \]

### Final Answer:
The acceleration of the particle is:
\[ \boxed{2a_2} \]