# Question #fca8b

Jan 22, 2017

The displacement ($x$) of the particle with time ($t$) along x-axis is given here by the following relation

$x = 2 - 5 t + 6 {t}^{2}$

Differntiating the relation w.r to we get the velocity of the particle at tth instant. So

$v \left(t\right) = \frac{\mathrm{dx}}{\mathrm{dt}} = 0 - 5 + 12 t$

Hence initially when $t = 0$ we have velocity
$v \left(0\right) = - 5 + 12 \times 0 = - 5 u n i t$