91Ó°ÊÓ

At time $t=0$, the initial position isrole="math" localid="1656991996634" ${\mathrm{x}}_0=-20\mathrm{m}$

The velocity is the time rate of change of displacement. The acceleration is the rate of change of velocity. The positive sign of velocity implies a positive change in displacement for the given time interval. The positive sign of acceleration implies a positive change in the velocity for the given time interval.

Formulae:

The final velocity is given by,

$v_f=v_0+at$

Here,role="math" localid="1656992492478" $v_f$is the final velocity,$v_0$is the initial velocity, is acceleration, and$t$ is time.

$v_f^2=v_0^2+2a∆x$

Here,$∆x$ is displacement.

The displacement is given by,

$∆x=v_{0}t+\frac{1}{2}a{t}^2$

The particle stops momentarily, so its final velocity would be zero. i.e. $v_{f=}0$

Hence,

$v_f=v_0+at$gives,

$0=v_0+at$

$t=\frac{-v_0}{a}$

Here, time would be positive when either of $v_0$or a will be negative.

This is possible only for situation 2 and 3.

Particle passes through the origin when$x_{\mathit{f}}\mathit{=}\mathit{0}$

Hence,

$$\begin{gathered} {\mathrm{x}}_{\mathrm{f}}={\mathrm{x}}_0+v_{0}t+\frac{1}{2}{\mathrm{at}}^2 \\ {\text{0=-20+v}}_{0}t+\frac{1}{2}{\mathrm{at}}^2 \end{gathered}$$

This gives quadratic equation in terms of t as,

$\frac{1}{2}a{t}^2+v_{0}t+-20=0$

We know that in

$a{x}^2+bx+c=0$

Root of equation will be positive when, $b^2-4ac=0$

Hence,$$\begin{gathered} v_0^2-4×\frac{1}{2}×a\left(-20\right)=0 \\ {v}_0^2=-40a \end{gathered}$$

This is possible only for situations 1 and 3.

Particle never passes through origin means it travels along negative x axis only.

And this is possible for situation 4 only.