If the displacement x of a particle (in metre) is related with time (in second) according to relation
`x = 2 t^3 - 3t^2 +2 t + 2`
find the position, velocity and acceleration of a particle at the end of 2 seconds.

Options

A. `10 m ; 14 m//s ; 10 m//s^2`
B. `18 m ; 14 m//s ; 18 m//s^2`
C. `14 m ; 112 m//s ; 18 m//s^2`
D. `10 m ; 14 m//s ; 18 m//s^2`

Explanation

To solve the problem step by step, we will find the position, velocity, and acceleration of the particle at the end of 2 seconds given the displacement function \( x(t) = 2t^3 - 3t^2 + 2t + 2 \).

### Step 1: Find the Position at \( t = 2 \) seconds
The position of the particle at any time \( t \) is given by the displacement function:
\[ x(t) = 2t^3 - 3t^2 + 2t + 2 \]
Now, we will substitute \( t = 2 \) into the equation:
\[ x(2) = 2(2^3) - 3(2^2) + 2(2) + 2 \]
Calculating each term:
\[ = 2(8) - 3(4) + 4 + 2 \]
\[ = 16 - 12 + 4 + 2 \]
\[ = 16 - 12 + 6 = 10 \text{ meters} \]

### Step 2: Find the Velocity
Velocity is the rate of change of displacement with respect to time, which is given by the derivative of the position function \( x(t) \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2t^3 - 3t^2 + 2t + 2) \]
Differentiating term by term:
\[ v(t) = 6t^2 - 6t + 2 \]
Now, we will substitute \( t = 2 \):
\[ v(2) = 6(2^2) - 6(2) + 2 \]
Calculating:
\[ = 6(4) - 12 + 2 \]
\[ = 24 - 12 + 2 = 14 \text{ meters per second} \]

### Step 3: Find the Acceleration
Acceleration is the rate of change of velocity with respect to time, which is given by the derivative of the velocity function \( v(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 6t + 2) \]
Differentiating term by term:
\[ a(t) = 12t - 6 \]
Now, we will substitute \( t = 2 \):
\[ a(2) = 12(2) - 6 \]
Calculating:
\[ = 24 - 6 = 18 \text{ meters per second squared} \]

### Summary of Results
- Position at \( t = 2 \) seconds: \( x = 10 \) meters
- Velocity at \( t = 2 \) seconds: \( v = 14 \) meters per second
- Acceleration at \( t = 2 \) seconds: \( a = 18 \) meters per second squared