Question

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?

Short answer

(a) At $2$the spot is instantaneously at rest.

(b) The value of $x$is$12cm$ when the spot is instantaneously at rest.

(c) The value of acceleration is$-9\mathrm{m}/{\mathrm{s}}^2$ when the spot is instantaneously at rest.

(d) The particle is movingright when it is coming to rest.

(e) The particle is moving to the left, just after coming to rest.

(f)At the time$3.46s$, the spot will reach an edge of the screen.

Step-by-step solution

Given data:

The position of the spot is given by the equation,

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

Relation between displacement, velocity, and acceleration

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.

(a) Calculation for a time after starting is the spot instantaneously at rest

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.

Step4:(b) Calculations for value of x when the spot instantaneously comes to rest

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.

Step 5:(c) Calculations for spot’s acceleration when the it 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

(d) Determination of direction of the spot before coming to rest

$V>0$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.

(e) Determination of direction of the spot after 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.

Step 8:(f) Calculation of time when it reaches an edge of the screen

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.