The position x of a particle varies with time t as `x=a t^(2)-b`. For what value of t acceleration is zero?

Options

A. `(2a)/(3b)`
B. `a//b`
C. `(a)/(3b)`
D. never

Explanation

To solve the problem, we need to find the time \( t \) at which the acceleration of the particle is zero, given the position function \( x = a t^2 - b \).

### Step-by-step Solution:

1. **Identify the Position Function**:
The position of the particle is given by the equation:
\[ x = a t^2 - b \]

2. **Differentiate to Find Velocity**:
The velocity \( v \) is the first derivative of the position \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(a t^2 - b) \]
Since \( a \) is a constant and \( b \) is also a constant, we differentiate:
\[ v = 2a t \]

3. **Differentiate to Find Acceleration**:
The acceleration \( a \) is the derivative of the velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(2a t) \]
Again, since \( 2a \) is a constant, we differentiate:
\[ a = 2a \]

4. **Determine When Acceleration is Zero**:
From the calculation, we see that the acceleration \( a = 2a \) is a constant and does not depend on time \( t \). Therefore, the acceleration is never zero unless \( a = 0 \).

5. **Conclusion**:
The acceleration is independent of time \( t \) and is equal to \( 2a \). Thus, there is no specific value of \( t \) for which the acceleration is zero unless \( a \) itself is zero.

### Final Answer:
The acceleration is zero only if \( a = 0 \); otherwise, it is constant and not dependent on \( t \).