If impulse/ varies with time t as f `(kg ms^(-1))=20t^(2)-40t`.The change in momentum is minimum at

Options

A. t=2s
B. t=1s
C. `t=(1)/(2)s`
D. `t=(3)/(2)s`

Explanation

To solve the problem of finding the time at which the change in momentum is minimum, we will follow these steps:

### Step 1: Understand the relationship between impulse and momentum
Impulse (I) is defined as the change in momentum (Δp) and is given by the integral of force (F) over time (t):
\[ I = \int F \, dt = \Delta p \]

### Step 2: Given the expression for impulse
The impulse is given as:
\[ I(t) = 20t^2 - 40t \]

### Step 3: Set up the function for change in momentum
We can denote the change in momentum as:
\[ \Delta p = 20t^2 - 40t \]

### Step 4: Differentiate the change in momentum with respect to time
To find the minimum change in momentum, we need to differentiate Δp with respect to t:
\[ \frac{d(\Delta p)}{dt} = \frac{d}{dt}(20t^2 - 40t) \]
Using the power rule of differentiation:
\[ \frac{d(\Delta p)}{dt} = 40t - 40 \]

### Step 5: Set the derivative equal to zero to find critical points
To find the time at which the change in momentum is minimum, we set the derivative equal to zero:
\[ 40t - 40 = 0 \]
Solving for t:
\[ 40t = 40 \implies t = 1 \text{ second} \]

### Step 6: Confirm that this point is a minimum using the second derivative test
Next, we need to check if this critical point is indeed a minimum by taking the second derivative of Δp:
\[ \frac{d^2(\Delta p)}{dt^2} = \frac{d}{dt}(40t - 40) \]
The second derivative is:
\[ \frac{d^2(\Delta p)}{dt^2} = 40 \]
Since the second derivative is positive (40 > 0), it confirms that the function has a minimum at \( t = 1 \) second.

### Conclusion
The change in momentum is minimum at \( t = 1 \) second.

---