For motion of an object along the x-axis the velocity v depends on the displacement x as `v=3x^(2)-2x`, then what is the acceleratiobn at `x=2m`.

Options

A. `48ms^(-2)`
B. `80ms^(-2)`
C. `18ms^(-2)`
D. `10ms^(-2)`

Explanation

To find the acceleration of the object at \( x = 2 \, \text{m} \) given the velocity function \( v = 3x^2 - 2x \), we can follow these steps:

### Step 1: Write down the given velocity function
The velocity \( v \) as a function of displacement \( x \) is given by:
\[ v = 3x^2 - 2x \]

### Step 2: Use the relationship between acceleration and velocity
The acceleration \( a \) can be expressed in terms of velocity \( v \) and displacement \( x \) using the formula:
\[ a = v \frac{dv}{dx} \]
Here, \( \frac{dv}{dx} \) is the derivative of the velocity with respect to displacement.

### Step 3: Calculate \( \frac{dv}{dx} \)
To find \( \frac{dv}{dx} \), we differentiate the velocity function \( v \):
\[ \frac{dv}{dx} = \frac{d}{dx}(3x^2 - 2x) = 6x - 2 \]

### Step 4: Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula
Now, substituting \( v \) and \( \frac{dv}{dx} \) into the acceleration formula:
\[ a = v \frac{dv}{dx} = (3x^2 - 2x)(6x - 2) \]

### Step 5: Simplify the expression for acceleration
Now we simplify the expression:
\[ a = (3x^2 - 2x)(6x - 2) \]
Expanding this:
\[ a = 3x^2 \cdot 6x - 3x^2 \cdot 2 - 2x \cdot 6x + 2x \cdot 2 \]
\[ = 18x^3 - 6x^2 - 12x^2 + 4x \]
\[ = 18x^3 - 18x^2 + 4x \]

### Step 6: Substitute \( x = 2 \, \text{m} \) into the acceleration expression
Now, we need to find the acceleration at \( x = 2 \, \text{m} \):
\[ a = 18(2)^3 - 18(2)^2 + 4(2) \]
Calculating each term:
\[ = 18(8) - 18(4) + 8 \]
\[ = 144 - 72 + 8 \]
\[ = 80 \, \text{m/s}^2 \]

### Final Answer
The acceleration at \( x = 2 \, \text{m} \) is:
\[ \boxed{80 \, \text{m/s}^2} \]