A particle is moving in a straight line. Its displacement at time `t` is given by `s(I n m)=4t^(2)+2t`, then its velocity and acceleration at time `t=(1)/(2)` second are

Options

A. `-2m//s,-8m//s^(2)`
B. `2m//s,6m//s^(2)`
C. `-2m//s,8m//s^(2)`
D. `2m//s,8m//s^(2)`

Explanation

To solve the problem, we need to find the velocity and acceleration of a particle whose displacement \( s \) is given by the equation:

\[ s(t) = 4t^2 + 2t \]

at time \( t = \frac{1}{2} \) seconds.

### Step 1: Find the Velocity

The velocity \( v \) of the particle is defined as the rate of change of displacement with respect to time. Mathematically, this is expressed as:

\[ v(t) = \frac{ds}{dt} \]

To find the velocity, we differentiate the displacement function \( s(t) \):

\[ v(t) = \frac{d}{dt}(4t^2 + 2t) \]

Using the power rule of differentiation:

\[ \frac{d}{dt}(t^n) = n t^{n-1} \]

we differentiate each term:

1. For \( 4t^2 \), the derivative is \( 8t \).
2. For \( 2t \), the derivative is \( 2 \).

Thus, we have:

\[ v(t) = 8t + 2 \]

### Step 2: Calculate Velocity at \( t = \frac{1}{2} \)

Now, we substitute \( t = \frac{1}{2} \) into the velocity equation:

\[ v\left(\frac{1}{2}\right) = 8\left(\frac{1}{2}\right) + 2 \]

Calculating this gives:

\[ v\left(\frac{1}{2}\right) = 4 + 2 = 6 \, \text{m/s} \]

### Step 3: Find the Acceleration

Acceleration \( a \) is defined as the rate of change of velocity with respect to time:

\[ a(t) = \frac{dv}{dt} \]

We need to differentiate the velocity function \( v(t) \):

\[ a(t) = \frac{d}{dt}(8t + 2) \]

Differentiating gives:

1. For \( 8t \), the derivative is \( 8 \).
2. For \( 2 \), the derivative is \( 0 \).

Thus, we have:

\[ a(t) = 8 \]

### Step 4: Conclusion

The acceleration is constant, and it does not depend on time. Therefore, at \( t = \frac{1}{2} \):

\[ a\left(\frac{1}{2}\right) = 8 \, \text{m/s}^2 \]

### Final Results

- Velocity at \( t = \frac{1}{2} \) seconds: \( 6 \, \text{m/s} \)
- Acceleration at \( t = \frac{1}{2} \) seconds: \( 8 \, \text{m/s}^2 \)