A particle moves along the $x-$axis in such a way that its $x-$ coordinates varies with time as $x=1–2t+3t^2$. What will be its initial velocity?

The correct option is C $-2\text{m/s}$
Here displacement, $x=1-2t+3t^2$
$v=\frac{dx}{dt}=\frac{d}{dt}(1-2t+3t^2)\Rightarrow v=-2+6t$
This is the velocity at time $t$, Initially $t=0$,
$\therefore v=-2\text{m/s}$
Hence, the initial velocity of the particle is $-2\text{m/s}$