Finding velocity of a particle in linear motion.

Question

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?

Answer 1

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;

$v(-1.03) = (-1.03-1)(-1.03-2)= 6.1509$

Answer 2

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 $$ T = \frac{1}{2} m \dot{x}^2 \approx 36 $$