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

The correct option is C

$–2,-8$

Step 1: Given data

Displacement $s=–4t^2+2t$

Step 2: Find velocity, $v$

Velocity is the rate of change of displacement with respect to the time.

$$\begin{gathered} v=\frac{ds}{dt} \\ v=\frac{d}{dt}\left(-4t^2+2t\right) \\ v=-8t+2 \\ {v}_{t=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=-8\times \frac{1}{2}+2 \\ {v}_{t=\frac{1}{2}}=-2 \end{gathered}$$ [ $v_{t=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is velocity at time $t=\frac{1}{2}$]

Step 3: Find acceleration, $a$

Acceleration is the rate of change of the velocity with respect to the time.

$$\begin{gathered} a=\frac{dv}{dt} \\ a=\frac{d}{dt}\left(-8t+2\right) \\ a=-8 \end{gathered}$$

Hence, option $\left(C\right)$ is correct.