Finding velocity of a particle in linear motion. 
In a kinematics question of linear motion I have been given a function between velocity$(v)$ and time$(t)$.
  $v=(t-1)(t-2)$
  and that for $t=0$, position of particle is $+4$.
  Now I have to find its velocity at origin.   

I integrated $v$ and got a cubic equation. Now I'm unable to solve. Is there any  way to solve this problem?
 A: You need to find for what $t$ the position is $0$.
integrating $v$ gives us position;
$s = \displaystyle\int(t-1)(t-2)\,dt = \int (t^2-3t+2)\,dt = \frac{t^3}3-\frac32t^2+2t+C$
given that $s(0) = 4$
$\implies 4 = C$
$\therefore s(t) = \displaystyle\frac{t^3}3-\frac32t^2+2t+4$
To find at what $t$ position is zero set $s(t) =0$
$s(t) =0 =\displaystyle\frac{t^3}3-\frac32t^2+2t+4$
The above is not easily solvable and you can find the solution using numerical methods . The only real solution is $x \approx -1.03$
So the velocity at origin is given by;
$v(-1.03) = (-1.03-1)(-1.03-2)= 6.1509$ 
A: You have
$$
\dot{x}(t) = (t-1)(t-2) \\
x(0) = 4
$$
This gives
\begin{align}
4 - x(t) 
&= \int\limits_{t}^0 (\tau-1)(\tau-2)\, d\tau \\
&= [(1/3)(\tau-1)^3-(1/2)(\tau-1)^2]_{\tau=t}^{\tau=0} \\
&= (1/3)(-1)-(1/2)-(1/3)(t-1)^3+(1/2)(t-1)^2
\end{align}
so
$$
x(t) = (1/3)(t-1)^3-(1/2)(t-1)^2+29/6
$$

(Large version)
We have
$$
x(-1) = (1/3)(-8)-(1/2)4+29/6 = (-16-12+29)/6 = 1/6 \approx 0
$$
and
$$
\dot{x}(-1)=6
$$
With $m=2$ this estimates the kinetic energy at $x=0$ as
$$
T = \frac{1}{2} m \dot{x}^2 \approx 36
$$
