A particle is moving along straight line whose position `x` at time `t` is described by `x = t^(3) - t^(2)` where `x` is in meters and `t`is in seconds . Then the average acceleration from `t = 2` sec. to `t = 4` sec, is :

Options

A. `16 m//s^(2)`
B. `18 m//s^(2)`
C. `22 m//s^(2)`
D. `10 m//s^(2)`

Explanation

To find the average acceleration of the particle from \( t = 2 \) seconds to \( t = 4 \) seconds, we will follow these steps:

### Step 1: Determine the position function
The position of the particle is given by the equation:
\[ x(t) = t^3 - t^2 \]

### Step 2: Find the velocity function
The velocity \( v(t) \) is the derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - t^2) = 3t^2 - 2t \]

### Step 3: Calculate the velocity at \( t = 4 \) seconds
Substituting \( t = 4 \) into the velocity function:
\[ v(4) = 3(4^2) - 2(4) = 3(16) - 8 = 48 - 8 = 40 \, \text{m/s} \]

### Step 4: Calculate the velocity at \( t = 2 \) seconds
Substituting \( t = 2 \) into the velocity function:
\[ v(2) = 3(2^2) - 2(2) = 3(4) - 4 = 12 - 4 = 8 \, \text{m/s} \]

### Step 5: Calculate the average acceleration
The average acceleration \( a_{avg} \) is given by the formula:
\[ a_{avg} = \frac{v_f - v_i}{\Delta t} \]
where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( \Delta t \) is the change in time.

Substituting the values we found:
- \( v_f = 40 \, \text{m/s} \)
- \( v_i = 8 \, \text{m/s} \)
- \( \Delta t = 4 - 2 = 2 \, \text{s} \)

\[ a_{avg} = \frac{40 - 8}{2} = \frac{32}{2} = 16 \, \text{m/s}^2 \]

### Final Answer
The average acceleration from \( t = 2 \) seconds to \( t = 4 \) seconds is:
\[ \boxed{16 \, \text{m/s}^2} \]