The position `x` of a particle with respect to time `t` along the x-axis is given by `x=9t^(2)-t^(3)` where `x` is in meter and `t` in second. What will be the position of this particle when it achieves maximum speed along the positive `x` direction

Options

A. 32 m
B. 54 m
C. 81 m
D. 24 m

Explanation

To find the position of the particle when it achieves maximum speed along the positive x direction, we can follow these steps:

### Step 1: Write down the position function
The position of the particle is given by the equation:
\[ x(t) = 9t^2 - t^3 \]

### Step 2: Differentiate to find the velocity
To find the velocity, we differentiate the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(9t^2 - t^3) \]
Using the power rule of differentiation:
\[ v(t) = 18t - 3t^2 \]

### Step 3: Differentiate to find the acceleration
Next, we differentiate the velocity function to find the acceleration:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(18t - 3t^2) \]
Again, using the power rule:
\[ a(t) = 18 - 6t \]

### Step 4: Set the acceleration to zero to find maximum speed
To find the time at which the particle achieves maximum speed, we set the acceleration equal to zero:
\[ 18 - 6t = 0 \]
Solving for \( t \):
\[ 6t = 18 \]
\[ t = 3 \, \text{seconds} \]

### Step 5: Substitute \( t \) back into the position function
Now we substitute \( t = 3 \) seconds back into the position function to find the position at that time:
\[ x(3) = 9(3^2) - (3^3) \]
Calculating this:
\[ x(3) = 9(9) - 27 \]
\[ x(3) = 81 - 27 \]
\[ x(3) = 54 \, \text{meters} \]

### Final Answer
The position of the particle when it achieves maximum speed along the positive x direction is:
\[ \boxed{54 \, \text{meters}} \]

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