A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by `(4t^3-2t)`, where t is in sec and velocity in m/s. what is the acceleration of the particle when it is 2 m from the origin?

Options

A. `28 m//s^(2) `
B. ` 22 m//s^(2) `
C. ` 12 m//s^(2) `
D. ` 10 m//s^(2)`

Explanation

To solve the problem step by step, we need to find the acceleration of a particle when it is 2 meters from the origin, given its velocity as a function of time.

### Step 1: Write down the velocity function
The velocity \( v(t) \) of the particle is given by:
\[ v(t) = 4t^3 - 2t \]

### Step 2: Relate velocity to displacement
Velocity is the derivative of displacement \( x \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} \]
Thus, we can write:
\[ dx = (4t^3 - 2t) dt \]

### Step 3: Integrate to find displacement
To find the displacement \( x \), we integrate the velocity function:
\[ x = \int (4t^3 - 2t) dt \]
Calculating the integral:
\[ x = \left( \frac{4}{4} t^4 - \frac{2}{2} t^2 \right) + C = t^4 - t^2 + C \]
Since the body starts from the origin, we set \( C = 0 \):
\[ x = t^4 - t^2 \]

### Step 4: Set displacement equal to 2 meters
We need to find the time \( t \) when the displacement \( x \) is 2 meters:
\[ t^4 - t^2 = 2 \]
Rearranging gives:
\[ t^4 - t^2 - 2 = 0 \]
Let \( u = t^2 \). Then the equation becomes:
\[ u^2 - u - 2 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \]
This gives us:
\[ u = 2 \quad \text{or} \quad u = -1 \quad (\text{not possible since } u = t^2) \]
Thus, \( t^2 = 2 \) implies:
\[ t = \sqrt{2} \text{ seconds} \]

### Step 6: Find acceleration
Acceleration \( a(t) \) is the derivative of velocity:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(4t^3 - 2t) = 12t^2 - 2 \]
Now, substituting \( t = \sqrt{2} \):
\[ a(\sqrt{2}) = 12(\sqrt{2})^2 - 2 = 12 \cdot 2 - 2 = 24 - 2 = 22 \text{ m/s}^2 \]

### Final Answer
The acceleration of the particle when it is 2 meters from the origin is:
\[ \boxed{22 \text{ m/s}^2} \]