The distance covered by a particle in t sec is given by `x=3+8t-4t^(2)`. What will be its velocity after 1 second.

Answer

To find the velocity of the particle after 1 second, we will follow these steps:

### Step 1: Write down the distance function
The distance covered by the particle in \( t \) seconds is given by the equation:
\[ x(t) = 3 + 8t - 4t^2 \]

### Step 2: Differentiate the distance function to find velocity
To find the velocity \( v(t) \), we need to take the derivative of the distance function \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(3 + 8t - 4t^2) \]
Using the power rule of differentiation:
- The derivative of a constant (3) is 0.
- The derivative of \( 8t \) is 8.
- The derivative of \( -4t^2 \) is \( -8t \).

Thus, we have:
\[ v(t) = 0 + 8 - 8t = 8 - 8t \]

### Step 3: Calculate the velocity at \( t = 1 \) second
Now, we need to find the velocity after 1 second by substituting \( t = 1 \) into the velocity function:
\[ v(1) = 8 - 8(1) = 8 - 8 = 0 \]

### Final Answer
The velocity of the particle after 1 second is:
\[ \boxed{0} \]