A particle moves along x-axis and its displacement at any time is given by `x(t) = 2t^(3) -3t^(2) + 4t` in SI units. The velocity of the particle when its acceleration is zero is

Options

A. `2.5 ms^(-1) `
B. `3.5 ms^(-1)`
C. `4.5 ms^(-1)`
D. `8.5 ms^(-1)`

Explanation

To solve the problem, we need to find the velocity of the particle when its acceleration is zero. The displacement of the particle is given by the function:

\[ x(t) = 2t^3 - 3t^2 + 4t \]

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

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2t^3 - 3t^2 + 4t) \]

Differentiating each term:

- The derivative of \( 2t^3 \) is \( 6t^2 \)
- The derivative of \( -3t^2 \) is \( -6t \)
- The derivative of \( 4t \) is \( 4 \)

Thus, the velocity function is:

\[ v(t) = 6t^2 - 6t + 4 \]

### Step 2: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity \( v(t) \) with respect to time \( t \):

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 6t + 4) \]

Differentiating each term:

- The derivative of \( 6t^2 \) is \( 12t \)
- The derivative of \( -6t \) is \( -6 \)
- The derivative of \( 4 \) is \( 0 \)

Thus, the acceleration function is:

\[ a(t) = 12t - 6 \]

### Step 3: Set acceleration to zero and solve for \( t \)
To find the time when acceleration is zero, we set \( a(t) \) to zero:

\[ 12t - 6 = 0 \]

Solving for \( t \):

\[ 12t = 6 \implies t = \frac{6}{12} = \frac{1}{2} \text{ seconds} \]

### Step 4: Find the velocity at \( t = \frac{1}{2} \)
Now, we need to find the velocity at \( t = \frac{1}{2} \) seconds by substituting \( t \) into the velocity function:

\[ v\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^2 - 6\left(\frac{1}{2}\right) + 4 \]

Calculating each term:

- \( 6\left(\frac{1}{2}\right)^2 = 6 \cdot \frac{1}{4} = \frac{6}{4} = \frac{3}{2} \)
- \( -6\left(\frac{1}{2}\right) = -3 \)
- The constant term is \( 4 \)

Combining these:

\[ v\left(\frac{1}{2}\right) = \frac{3}{2} - 3 + 4 = \frac{3}{2} - \frac{6}{2} + \frac{8}{2} = \frac{3 - 6 + 8}{2} = \frac{5}{2} \]

Thus, the velocity when acceleration is zero is:

\[ v\left(\frac{1}{2}\right) = \frac{5}{2} \text{ m/s} = 2.5 \text{ m/s} \]

### Final Answer
The velocity of the particle when its acceleration is zero is **2.5 m/s**.

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