Displacement of a particle moving in a straight line is written as follows:
`x=(t^(3))/(3)-(5t^(2))/(2)+6t+7`
what is the possible acceleration of particle when particle is in a state of rest?

Options

A. `1m//s^(2)`
B. `-1m//s^(2)`
C. `-5m//s^(2)`
D. `+5m//s^(2)`

Explanation

To find the possible acceleration of a particle when it is at rest, we will follow these steps:

### Step 1: Write down the displacement equation
The displacement of the particle is given by:
\[ x = \frac{t^3}{3} - \frac{5t^2}{2} + 6t + 7 \]

### Step 2: Find the velocity
Velocity \( v \) is the first derivative of displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{t^3}{3} - \frac{5t^2}{2} + 6t + 7 \right) \]
Calculating the derivative:
\[ v = t^2 - 5t + 6 \]

### Step 3: Set the velocity to zero
To find when the particle is at rest, we set the velocity \( v \) to zero:
\[ t^2 - 5t + 6 = 0 \]

### Step 4: Solve the quadratic equation
We can factor the quadratic equation:
\[ (t - 2)(t - 3) = 0 \]
Thus, the solutions for \( t \) are:
\[ t = 2 \quad \text{and} \quad t = 3 \]

### Step 5: Find the acceleration
Acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(t^2 - 5t + 6) \]
Calculating the derivative:
\[ a = 2t - 5 \]

### Step 6: Calculate acceleration at \( t = 2 \) and \( t = 3 \)
1. For \( t = 2 \):
\[ a = 2(2) - 5 = 4 - 5 = -1 \, \text{m/s}^2 \]

2. For \( t = 3 \):
\[ a = 2(3) - 5 = 6 - 5 = 1 \, \text{m/s}^2 \]

### Conclusion
The possible accelerations of the particle when it is at rest are:
- At \( t = 2 \): \( -1 \, \text{m/s}^2 \)
- At \( t = 3 \): \( 1 \, \text{m/s}^2 \)