Force applied on a body is given by ltbygt `F=(3t^(2)-2t+10)`N where t is in seconds. Find impulse imparted in t=0 to t=2 sec.

Answer

To find the impulse imparted on a body when a force is applied, we can use the formula for impulse, which is given by the integral of force over time. The force is provided as a function of time, \( F(t) = 3t^2 - 2t + 10 \) N. We need to calculate the impulse from \( t = 0 \) to \( t = 2 \) seconds.

### Step-by-Step Solution:

1. **Write the formula for impulse**:
\[ \text{Impulse} = \int_{t_1}^{t_2} F(t) \, dt \]
where \( t_1 = 0 \) s and \( t_2 = 2 \) s.

2. **Substitute the force function into the integral**:
\[ \text{Impulse} = \int_{0}^{2} (3t^2 - 2t + 10) \, dt \]

3. **Integrate the function**:
- The integral of \( 3t^2 \) is \( t^3 \).
- The integral of \( -2t \) is \( -t^2 \).
- The integral of \( 10 \) is \( 10t \).

Thus, we can write:
\[ \int (3t^2 - 2t + 10) \, dt = t^3 - t^2 + 10t + C \]

4. **Evaluate the definite integral from 0 to 2**:
\[ \text{Impulse} = \left[ t^3 - t^2 + 10t \right]_{0}^{2} \]

5. **Calculate the upper limit (t = 2)**:
\[ = (2^3 - 2^2 + 10 \cdot 2) = (8 - 4 + 20) = 24 \]

6. **Calculate the lower limit (t = 0)**:
\[ = (0^3 - 0^2 + 10 \cdot 0) = 0 \]

7. **Subtract the lower limit from the upper limit**:
\[ \text{Impulse} = 24 - 0 = 24 \, \text{N·s} \]

### Final Answer:
The impulse imparted in the time interval from \( t = 0 \) to \( t = 2 \) seconds is \( 24 \, \text{N·s} \).

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