The linear momentum P of a particle varies with time as follows `P=a+bt^(2)`
Where a and b are constants. The net force acting on the particle is

Options

A. proportional to t
B. proportional to `t^(2)`
C. Zero
D. constant

Answer

To find the net force acting on a particle whose linear momentum \( P \) varies with time as \( P = a + bt^2 \), we can use the relationship between force and momentum. According to Newton's second law, the net force \( F \) acting on an object is equal to the rate of change of momentum with respect to time:

\[ F = \frac{dP}{dt} \]

### Step 1: Differentiate the momentum with respect to time
Given the momentum function:

\[ P = a + bt^2 \]

we differentiate \( P \) with respect to \( t \):

\[ \frac{dP}{dt} = \frac{d}{dt}(a + bt^2) \]

### Step 2: Apply the differentiation rules
Since \( a \) is a constant, its derivative is zero. For the term \( bt^2 \), we apply the power rule of differentiation:

\[ \frac{d}{dt}(bt^2) = b \cdot \frac{d}{dt}(t^2) = b \cdot 2t = 2bt \]

### Step 3: Write the expression for net force
Now, substituting back into the equation for force, we have:

\[ F = \frac{dP}{dt} = 0 + 2bt = 2bt \]

### Conclusion
Thus, the net force acting on the particle is:

\[ F = 2bt \]

This shows that the net force is directly proportional to time \( t \).