Velocity of particle moving along x-axis is given as `v = ( x^(3) - x^(2) + 2)` m`//`sec. Find the acceleration of particle at x=2 meter.

Options

A. `48 m//s^(2)`
B. `16 m//s^(2)`
C. `32 m//s^(2)`
D. `8 m//s^(2)`

Answer

To find the acceleration of a particle moving along the x-axis with a given velocity function, we can follow these steps:

### Step 1: Write down the given velocity function
The velocity \( v \) of the particle is given by:
\[ v = x^3 - x^2 + 2 \quad \text{(in m/s)} \]

### Step 2: Understand the relationship between acceleration and velocity
Acceleration \( a \) is defined as the rate of change of velocity with respect to time, which can be expressed as:
\[ a = \frac{dv}{dt} \]
However, since the velocity is given as a function of position \( x \), we can use the chain rule to express acceleration as:
\[ a = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Here, \( \frac{dx}{dt} \) is the velocity \( v \).

### Step 3: Differentiate the velocity function with respect to \( x \)
We need to find \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = \frac{d}{dx}(x^3 - x^2 + 2) \]
Calculating the derivative:
\[ \frac{dv}{dx} = 3x^2 - 2x \]

### Step 4: Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula
Now we can express acceleration as:
\[ a = v \cdot \frac{dv}{dx} \]
Substituting the expressions we have:
\[ a = (x^3 - x^2 + 2)(3x^2 - 2x) \]

### Step 5: Evaluate the acceleration at \( x = 2 \) meters
Substituting \( x = 2 \) into the expressions for \( v \) and \( \frac{dv}{dx} \):
1. Calculate \( v \):
\[ v = 2^3 - 2^2 + 2 = 8 - 4 + 2 = 6 \, \text{m/s} \]
2. Calculate \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = 3(2^2) - 2(2) = 3(4) - 4 = 12 - 4 = 8 \]
3. Now substitute these values into the acceleration formula:
\[ a = 6 \cdot 8 = 48 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the particle at \( x = 2 \) meters is:
\[ \boxed{48 \, \text{m/s}^2} \]