Kinematics

Let the velocity at time $t$ be $v(t)$ .

Then, $v(t) = \int a(t) \ dt = \int (12t^3 - 2) \ dt = 3t^4 - 2t + C$ , where $C$ is the constant of integration.

It is given that:

$\begin{aligned} v(6.0) & = -6.0 \\ \implies 3(6.0)^4 - 2(6.0) + C & = -6.0 \\ \implies C & = - 3882.0 \\ \implies v(t) & = 3t^4 - 2t - 3882.0 \\ v(0) & = \boxed{- 3882.0} \text{ m/s} \end{aligned}$ .

Setting v(6.0s)=−6 m/s , we can find the unknown constant C

v(t)= 3t^4 -2t +C ................................... v(6.0)=−6.0=3×(6.0)4−2×(6.0)+C⟹C=−3882.0 m/s

For time t=0, v(0)=C=−3882.0 m/s.