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. We start with the given position function:

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

### Step 1: Find the Velocity
The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):

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

Using the power rule for differentiation, we differentiate each term:

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

### Step 2: Find the Time at Maximum Speed
To find the time when the speed is maximum, we need to set the derivative of the velocity (the acceleration) to zero:

\[ \frac{dv}{dt} = 0 \]

First, we differentiate the velocity function:

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

Setting this equal to zero gives:

\[ 24 - 12t = 0 \]
\[ 12t = 24 \]
\[ t = 2 \text{ seconds} \]

### Step 3: Verify Maximum Speed Condition
To ensure that this time corresponds to a maximum speed, we check the second derivative of the velocity:

\[ \frac{d^2v}{dt^2} = -12 \]

Since \(-12 < 0\), this confirms that the speed is indeed at a maximum at \( t = 2 \) seconds.

### Step 4: Find the Position at Maximum Speed
Now, we substitute \( t = 2 \) seconds back into the original position function to find the position:

\[ y(2) = 12(2^2) - 2(2^3) \]
\[ = 12(4) - 2(8) \]
\[ = 48 - 16 \]
\[ = 32 \text{ meters} \]

### Final Answer
The position of the particle when it achieves maximum speed is \( 32 \) meters.

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