The position of a particle with respect to time t along y-axis is given by : `y = 12t^2 – 2t^3`, where, y is in metres and t is in seconds. When the particle achieves maximum speed, the position of the particle would be

Options

A. 64 m
B. 40 m
C. 16 m
D. 32 m

Answer

To solve the problem, we need to find the position of the particle when it achieves maximum speed. The position of the particle is given by the equation:

\[ y = 12t^2 - 2t^3 \]

### Step 1: Find the velocity
To find the speed of the particle, we need to differentiate the position function with respect to time \( t \).

\[ v = \frac{dy}{dt} = \frac{d}{dt}(12t^2 - 2t^3) \]

Using the power rule of differentiation:

\[ v = 24t - 6t^2 \]

### Step 2: Find the time at which speed is maximum
To find the time when the speed is maximum, we need to set the derivative of the velocity function to zero. This means we need to differentiate the velocity function \( v \) with respect to \( t \):

\[ \frac{dv}{dt} = \frac{d}{dt}(24t - 6t^2) \]

Differentiating gives:

\[ \frac{dv}{dt} = 24 - 12t \]

Setting this equal to zero to find the critical points:

\[ 24 - 12t = 0 \]

Solving for \( t \):

\[ 12t = 24 \implies t = 2 \, \text{seconds} \]

### Step 3: Verify that this is a maximum
To confirm that this point is a maximum, we can check the second derivative of the velocity:

\[ \frac{d^2v}{dt^2} = \frac{d}{dt}(24 - 12t) = -12 \]

Since the second derivative is negative, this confirms that the speed is indeed at a maximum when \( t = 2 \) seconds.

### Step 4: Find the position at maximum speed
Now we need to find the position of the particle at \( t = 2 \) seconds. We substitute \( t = 2 \) into the original position equation:

\[ y = 12(2^2) - 2(2^3) \]

Calculating this:

\[ y = 12(4) - 2(8) = 48 - 16 = 32 \, \text{meters} \]

### Final Answer
Thus, when the particle achieves maximum speed, the position of the particle is:

\[ \boxed{32 \, \text{meters}} \]