The position x of a particle varies with time t as `x=at^(2)-bt^(3)`.The acceleration of the particle will be zero at time t equal to

Options

A. `(a)/(b)`
B. `(2a)/(3b)`
C. `(a)/(3b)`
D. Zero

Answer

To solve the problem, we need to find the time \( t \) at which the acceleration of the particle is zero, given the position function \( x(t) = at^2 - bt^3 \).

### Step-by-Step Solution:

1. **Write the position function:**
\[ x(t) = at^2 - bt^3 \]

2. **Find the velocity function:**
The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(at^2 - bt^3) \]
Using the power rule for differentiation:
\[ v(t) = 2at - 3bt^2 \]

3. **Find the acceleration function:**
The acceleration \( a(t) \) is the derivative of the velocity function with respect to time \( t \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(2at - 3bt^2) \]
Again, applying the power rule:
\[ a(t) = 2a - 6bt \]

4. **Set the acceleration to zero to find the time:**
We need to find the time \( t \) when the acceleration is zero:
\[ 2a - 6bt = 0 \]
Rearranging this equation gives:
\[ 6bt = 2a \]
Dividing both sides by \( 6b \):
\[ t = \frac{2a}{6b} = \frac{a}{3b} \]

5. **Final answer:**
The time at which the acceleration of the particle is zero is:
\[ t = \frac{a}{3b} \]