The position function x(t) of a particle moving along an xaxis is$\mathbf{x}\mathbf{=}\mathbf{4}\mathbf{-}\mathbf{6}{\mathbf{t}}^{\mathbf{2}}$, with xin metres and tin seconds. (a) At what time does the particle momentarily stop? (b) Where does the particle momentarily stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph x vs t for range -5 secto +5 sec. (f) To shift the curve rightward on the graph, should we include the term +20tor -20tin x(t)? (g) Does that inclusion increase or decrease the value of x at which the particle momentarily stops?

$x=4-6t^2$

The problem involves differentiation of the quantity. Here the functional notation can be used for differentiation. To initiate,vandacan be calculated from the given equation by taking its derivative. By plugging the value ofin equation of position to find position at this time.At x(t)=0 solve quadratic equation for t. The graph forx vst can be drawn. Also by adding 20t we can draw shifted graph.

Formula:

$v\left(t\right)=\frac{dx\left(t\right)}{dt}$

Find velocity and acceleration from given position by taking derivative of it

$$\begin{gathered} v\left(t\right)=\frac{dx\left(t\right)}{dt} \\ =-12t \\ a\left(t\right)=\frac{dv\left(t\right)}{dt} \\ =-12 \end{gathered}$$

It is found that at t=0 the velocity is 0.

$x=4-6t^2$

At t=0 the position of particle is$x\left(0\right)=4.0\mathrm{m}$

Solve$x\left(t\right)=0$

$0=4.0-6.0t^2$

By solving this quadratic equation for the negative time$t=-0.82\mathrm{s}$

Alsothe positive time from above quadratic equation$t=+0.82\mathrm{s}$

Both the graphs (on the left) as well as the shifted graph can be drawn. In both graphs time limits are$-3\le t\le 3$

The graph 2 can be drawn by adding 20t in$x\left(t\right)$expression.

The slopes of graphs as zero, shift causes $v=0$at larger values of x.