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 equation
The position \( x \) of the particle with respect to time \( t \) is given by:
\[ x = 9t^2 - t^3 \]

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

### Step 3: Find the maximum speed
To find when the speed is maximum, we need to find the critical points of the velocity function. We do this by setting the derivative of the velocity (which is the acceleration) to zero:
\[ \frac{dv}{dt} = 0 \]
Calculating the derivative of \( v \):
\[ \frac{dv}{dt} = 18 - 6t \]
Setting this equal to zero:
\[ 18 - 6t = 0 \]
Solving for \( t \):
\[ 6t = 18 \implies t = 3 \text{ seconds} \]

### Step 4: Find the position at maximum speed
Now that we know the time at which the speed is maximum, we can substitute \( t = 3 \) seconds back into the position equation to find the position:
\[ x = 9(3^2) - (3^3) \]
Calculating this:
\[ x = 9(9) - 27 = 81 - 27 = 54 \text{ meters} \]

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

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