A particle is moving with a velocity of `v=(3+ 6t +9t^2) m/s.` Find out
(a) the acceleration of the particle at `t=3 s.`
(b) the displacement of the particle in the interval `t=5s` to `t=8 s.`

Explanation

To solve the problem step by step, we will address both parts (a) and (b) systematically.

### Part (a): Find the acceleration of the particle at `t = 3 s`.

1. **Given Velocity Function**: The velocity of the particle is given by the equation:
\[ v(t) = 3 + 6t + 9t^2 \quad \text{(in m/s)} \]

2. **Acceleration Definition**: Acceleration \(a\) is defined as the rate of change of velocity with respect to time. Mathematically, this can be expressed as:
\[ a(t) = \frac{dv}{dt} \]

3. **Differentiate the Velocity Function**: We will differentiate the velocity function \(v(t)\) with respect to \(t\):
\[ a(t) = \frac{d}{dt}(3 + 6t + 9t^2) \]
- The derivative of \(3\) is \(0\).
- The derivative of \(6t\) is \(6\).
- The derivative of \(9t^2\) is \(18t\).

Thus, we have:
\[ a(t) = 0 + 6 + 18t = 6 + 18t \]

4. **Calculate Acceleration at \(t = 3 s\)**: Now, substitute \(t = 3\) into the acceleration equation:
\[ a(3) = 6 + 18(3) = 6 + 54 = 60 \quad \text{(in m/s}^2\text{)} \]

5. **Final Result for Part (a)**: The acceleration of the particle at \(t = 3 s\) is:
\[ a = 60 \, \text{m/s}^2 \]

### Part (b): Find the displacement of the particle in the interval from `t = 5 s` to `t = 8 s`.

1. **Displacement Definition**: Displacement \(x\) can be found by integrating the velocity function over the given time interval:
\[ x = \int_{t_1}^{t_2} v(t) \, dt \]
where \(t_1 = 5\) s and \(t_2 = 8\) s.

2. **Set Up the Integral**: Substitute the velocity function into the integral:
\[ x = \int_{5}^{8} (3 + 6t + 9t^2) \, dt \]

3. **Integrate Each Term**:
- The integral of \(3\) with respect to \(t\) is \(3t\).
- The integral of \(6t\) is \(3t^2\).
- The integral of \(9t^2\) is \(3t^3\).

Thus, we have:
\[ x = \left[ 3t + 3t^2 + 3t^3 \right]_{5}^{8} \]

4. **Evaluate the Integral at the Limits**:
- First, evaluate at \(t = 8\):
\[ x(8) = 3(8) + 3(8^2) + 3(8^3) = 24 + 192 + 1536 = 1752 \]
- Next, evaluate at \(t = 5\):
\[ x(5) = 3(5) + 3(5^2) + 3(5^3) = 15 + 75 + 375 = 465 \]

5. **Calculate the Displacement**:
\[ \Delta x = x(8) - x(5) = 1752 - 465 = 1287 \quad \text{(in meters)} \]

6. **Final Result for Part (b)**: The displacement of the particle from \(t = 5 s\) to \(t = 8 s\) is:
\[ \Delta x = 1287 \, \text{m} \]

### Summary of Results:
- (a) The acceleration at \(t = 3 s\) is \(60 \, \text{m/s}^2\).
- (b) The displacement from \(t = 5 s\) to \(t = 8 s\) is \(1287 \, \text{m}\).