A particle is acted upon by a force given by `F=(12t-3t^(2))`N, where is in seconds. Find the change in momenum of that particle from t=1 to t=3 sec.

Answer

To find the change in momentum of a particle acted upon by a force \( F = (12t - 3t^2) \) N from \( t = 1 \) to \( t = 3 \) seconds, we can follow these steps:

### Step 1: Understand the relationship between force and momentum
The change in momentum (\( \Delta p \)) of a particle is equal to the impulse applied to it, which can be calculated using the integral of force with respect to time:
\[ \Delta p = \int_{t_1}^{t_2} F \, dt \]

### Step 2: Set up the integral
Given the force \( F = 12t - 3t^2 \), we need to evaluate the integral from \( t = 1 \) to \( t = 3 \):
\[ \Delta p = \int_{1}^{3} (12t - 3t^2) \, dt \]

### Step 3: Integrate the force function
We can break this integral into two parts:
\[ \Delta p = \int_{1}^{3} 12t \, dt - \int_{1}^{3} 3t^2 \, dt \]

Now, we compute each integral separately.

1. For the first integral:
\[ \int 12t \, dt = 12 \cdot \frac{t^2}{2} = 6t^2 \]

2. For the second integral:
\[ \int 3t^2 \, dt = 3 \cdot \frac{t^3}{3} = t^3 \]

### Step 4: Evaluate the integrals at the limits
Now we evaluate both integrals from \( t = 1 \) to \( t = 3 \):
\[ \Delta p = \left[ 6t^2 \right]_{1}^{3} - \left[ t^3 \right]_{1}^{3} \]

Calculating each part:
1. For \( 6t^2 \):
\[ 6(3^2) - 6(1^2) = 6(9) - 6(1) = 54 - 6 = 48 \]

2. For \( t^3 \):
\[ (3^3) - (1^3) = 27 - 1 = 26 \]

### Step 5: Combine the results
Now, we combine the results:
\[ \Delta p = 48 - 26 = 22 \, \text{kg m/s} \]

### Final Answer
The change in momentum of the particle from \( t = 1 \) to \( t = 3 \) seconds is:
\[ \Delta p = 22 \, \text{kg m/s} \]

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