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

Options

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

Explanation

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

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

### Step 2: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(9t^2 - t^3) \]
Using the power rule:
\[ v(t) = 18t - 3t^2 \]

### Step 3: Find the maximum speed
To find the time at which the speed is maximum, we need to find when the acceleration is zero. The acceleration \( a(t) \) is the derivative of the velocity function:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(18t - 3t^2) \]
Using the power rule:
\[ a(t) = 18 - 6t \]

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

### Step 5: Find the position at \( t = 3 \, \text{seconds} \)
Now substitute \( t = 3 \) back into the position function to find the position:
\[ x(3) = 9(3^2) - (3^3) \]
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 +x direction is:
\[ \boxed{54 \, \text{meters}} \]