In an arcade video game, a spot is programmed to move across the screen according to$\mathbf{x}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{750}{\mathbf{t}}^{\mathbf{3}}$, where x is distance in centimeters measured from the left edge of the screen androle="math" localid="1656154621648" $\mathbf{t}$is time in seconds. When the spot reaches a screen edge, at either$\mathbf{x}\mathbf{=}\mathbf{0}$or$\mathbf{x}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{cm}$, t is reset to$\mathbf{0}$and the spot starts moving again according to$\mathbf{x}\left(t\right)$. (a) At what time after starting is the spot instantaneously at rest? (b) At what value of x does this occur? (c) What is the spot’s acceleration (including sign) when this occurs? (d)Is it moving right or left just prior to coming to rest? (e) Just after?(f) At what time$\mathbf{t}\mathbf{>}\mathbf{0}$does it first reach an edge of the screen?

The position of the spot is given by the equation,

$x=9.00t-0.750t^3$,

If you differentiate displacement and velocity equations, you can find the velocity and acceleration, respectively. Similarly, if you integrate acceleration and velocity equations, you can find velocity and displacement.

The particle is said to be at rest when $V=0$. It is also given that,

$x=9.00t-0.75t^3$

Differentiate the above equation with respect to t,

$$\begin{gathered} \frac{dx}{dt}=\frac{d\left(9.00t-0.75t^3\right)}{dt} \\ V=9.00-\left(0.75\times 2\right)t^2 \\ =9.00-2.25t^2 \end{gathered}$$

Now, put the value of $V=0$, to find $t$.

$$\begin{gathered} V=9.00-2.25t^2 \\ 0=9.00-2.25t^2 \end{gathered}$$

Solve this equation for t.

$$\begin{gathered} t^2=\frac{9.00}{2.25} \\ =4 \\ t=\sqrt{4} \\ =2s \end{gathered}$$

Hence, at $t=2\sec$the spot comes to rest instantaneously.

We have $t=2\sec$, when $V=0$, So the value for $x$will be,

$$\begin{gathered} x=9.00t-0.75t^3 \\ =9.00\times 2s\left(0.75\times {\left(2s\right)}^3\right) \\ =18-6 \\ =12\mathrm{cm} \end{gathered}$$

Hence, at $12$the spot comes to rest instantaneously.

It is given that,

$V=9.00-2.25t^2$

Differentiate above equation with respect to t,

$$\begin{gathered} \frac{dv}{dt}=\frac{d\left(9.00-2.25t^2\right)}{dt} \\ a=0-\left(2.25\times 2\right)t \\ =4.5t \end{gathered}$$

Also,$t=2sec$, therefore,

$$\begin{gathered} a=-4.5\times 2 \\ =-9m/s^2 \end{gathered}$$

Hence, at $a=9m/s^2$the spot comes to rest instantaneously

$V>0$for the time less than$t=2s$, i.e. before coming to rest.

From above statement we can conclude that particle is moving to right, when it is coming to rest.

The particle moves to the left immediately when it comes to rest.

From above statement we can conclude that particle is moving to left, just after coming to rest.

At the edge of the screen $x=0$,

$$\begin{gathered} x=9.00t-0.75t^3 \\ =9.00t-0.75t^3 \\ 0.75t^3=9.00t \\ 0.75t^2=9.00 \end{gathered}$$

Calculate the time t from the above equation.

$$\begin{gathered} t^2=\frac{9.00}{0.75} \\ =12 \\ t=\sqrt{12} \\ =3.46s \end{gathered}$$

Hence, at $t=3.46s$spot will reach the edge of the screen.