91Ó°ÊÓ

$\mathrm{x}=9.75+1.50{\mathrm{t}}^3$

The problem deals with the average velocity which is the displacement over the time. Also, it involves the instantaneous velocity of an object. It is the limit of an average velocity as the elapsed time approaches zero. Using a standard equation the average velocity in terms of displacement and time can be found. Further, instantaneous velocity can be found by taking derivative of equation of position.

Formula:

$v_{avg}=\frac{∆x}{∆t}$ (i)

${\mathrm{v}}_{\mathrm{instantaneous}}=\frac{\mathrm{dx}}{\mathrm{dt}}$ (ii)

It is given that,

$\mathrm{x}=9.75+1.50{\mathrm{t}}^3$

By plugging the values of t in the above equation we get,

$$\begin{gathered} {\mathrm{x}}^2=9.75+1.50{\left(2.00\right)}^3 \\ =21.75\mathrm{cm} \\ {\mathrm{x}}_3=9.75+1.50{\left(3.00\right)}^3 \\ =50.25\mathrm{cm} \end{gathered}$$

Using equation (i) the average velocity is,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{avg}}=\frac{∆\mathrm{x}}{∆\mathrm{t}} \\ =\frac{50.25-21.75}{3.00-2.00} \\ =28.5\mathrm{cm}/\mathrm{s} \end{gathered}$$

Average velocity is 28.5 cm/s

Using equation (ii) the instantaneous velocity is

$$\begin{gathered} v=\frac{dx}{dt} \\ =4.5t^2 \end{gathered}$$

At $$\begin{gathered} \mathrm{t}=2.00\mathrm{s} \\ \mathrm{v}=4.5{\left(2.00\right)}^2 \\ =18.0\mathrm{cm}/\mathrm{s} \end{gathered}$$

Instantaneous velocity at t=3.00 s,

$$\begin{gathered} \mathrm{v}=4.5{\left(3.00\right)}^2 \\ =40.5\mathrm{cm}/\mathrm{s} \end{gathered}$$

Instantaneous velocity at t=2.50 s

$$\begin{gathered} \mathrm{v}=4.5{\left(2.50\right)}^2 \\ =28.1\mathrm{cm}/\mathrm{s} \end{gathered}$$

Trying to find time of particle at midway between$x_2$ and $x_3$, we get,

$$\begin{gathered} {\mathrm{x}}_{\mathrm{m}}=\frac{50.25+21.75}{2} \\ =36\mathrm{cm} \\ {\mathrm{x}}_{\mathrm{m}}=9.75+1.50{\mathrm{t}}_{\mathrm{m}}^3 \\ 36=9.75+1.50{\mathrm{t}}_{\mathrm{m}}^3 \\ {\mathrm{t}}_{\mathrm{m}}=2.596\mathrm{s} \end{gathered}$$

Time at midway is 2.596 s

Thus, instantaneous speed at this time

$$\begin{gathered} \mathrm{v}=4.5\left(2.596\right)2 \\ =30.3\mathrm{cm}/\mathrm{s} \end{gathered}$$

Straight line slope between t=2s to 3s gives answer of part (a) and slope of tangent at the required points would give the answers of (b), (c), (d), (e)