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 moving along the X-axis, given its velocity function. The velocity is defined as \( v(t) = 4t^3 - 2t \).

### Step 1: Understand the relationship between displacement, velocity, and time.
The velocity \( v \) is the rate of change of displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]

### Step 2: Set up the equation for displacement.
We can express the displacement \( dx \) in terms of \( dt \):
\[ dx = (4t^3 - 2t) dt \]

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

### Step 4: Set the 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:
\[ u = 2 \quad \text{or} \quad u = -1 \]
Since \( u = t^2 \), we discard \( u = -1 \) and take \( u = 2 \):
\[ t^2 = 2 \implies t = \sqrt{2} \text{ seconds} \]

### Step 6: Find the acceleration at \( t = \sqrt{2} \).
The acceleration \( a \) is the derivative of the velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Given \( v(t) = 4t^3 - 2t \), we differentiate:
\[ a = \frac{d}{dt}(4t^3 - 2t) = 12t^2 - 2 \]

### Step 7: Substitute \( t = \sqrt{2} \) into the acceleration formula.
Now we substitute \( t = \sqrt{2} \):
\[ a = 12(\sqrt{2})^2 - 2 = 12 \cdot 2 - 2 = 24 - 2 = 22 \text{ m/s}^2 \]

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