A particle is moving in a straight line. At time t, the distance between the particle from its starting point is given by `x =t- 9t^(2) + t^(3)`. lts acceleration will be zero at

Options

A. t=1 unit time
B. t=2 unit time
C. t=3 unit time
D. t=4 unit time

Answer

To find when the acceleration of the particle is zero, we need to follow these steps:

### Step 1: Write the distance function
The distance \( x \) of the particle from its starting point is given by:
\[ x = t - 9t^2 + t^3 \]

### Step 2: Find the velocity
The velocity \( v \) is the first derivative of the distance \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(t - 9t^2 + t^3) \]
Differentiating term by term:
\[ v = 1 - 18t + 3t^2 \]

### Step 3: Find the acceleration
The acceleration \( a \) is the derivative of the velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(1 - 18t + 3t^2) \]
Differentiating term by term:
\[ a = 0 - 18 + 6t = 6t - 18 \]

### Step 4: Set the acceleration to zero
To find when the acceleration is zero, we set the equation for acceleration to zero:
\[ 6t - 18 = 0 \]

### Step 5: Solve for \( t \)
Solving for \( t \):
\[ 6t = 18 \\ t = \frac{18}{6} \\ t = 3 \]

### Conclusion
The acceleration of the particle will be zero at \( t = 3 \) units of time.

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