Speed or Velocity ??

As we have got a linear equation for v , we know its a constantly accelerated motion in a straight line. So using the general symbols.

$v=u+at$ we have $v=-6+2t$ , comparing these its clear that $u=-6$ and $a=2$ .

Now, knowing $x=ut+\frac {1}{2}at^2$ we substitute for the first second and find $\boxed{x=-5}$ . !!!!!!!!!! But as per the question $x=8$ so the $INITIAL POINT =13$ .

This is because in the first second it has travelled 5 places left of INITIAL POINT , AND we land on $+8$ .

Going in this way we find that when $t=4$ we get $x=-8$ .

Hence going 8 places left of 13 we get to THE FINAL POSITION is $\boxed{5}$

For further clarification comment and I will respond.

We know that to find velocity, we have to use the formula,

$\int^{T_2}_{T_1} v(t) = x(T_2)-x(T_1)$

so we substitute $T_2 = 1$ , $T_1=0$ and $v(t)=2t-6$ , we have

$\int^{1}_{0} 2t-6 = x(1)-x(0)$

$\left. t^2-6t \right|^1_0 = 8-x(0)$

$(1^2-6(1))-(0^2-6(0)) = 8-x(0)$

$\implies x(0) = 13$

Then, we do the same thing again, just change the value of $T_2$ become $4$ .

$\int^{4}_{0} 2t-6 = x(4)-x(0)$

$\left. t^2-6t \right|^4_0 = x(4)-13$

$(4^2-6(4))-(0^2-6(0)) = x(4)-13$

$\implies x(4) = \boxed{5}$