91Ó°ÊÓ

We are given with the equations:

$$\begin{gathered} 2.5\mathrm{rad}=0\mathrm{rad}+{\mathrm{Ó\neg }}_{\mathrm{i}}(10\mathrm{s})+\left(\right(1.5\mathrm{m}/{\mathrm{s}}^2)/2(50\mathrm{m}\left)\right){(10\mathrm{s})}^2 \\ {\mathrm{Ó\neg }}_{\mathrm{f}}={\mathrm{Ó\neg }}_{\mathrm{i}}+\left(\right(1.5\mathrm{m}/{\mathrm{s}}^2)/(50\mathrm{m}\left)\right)(10\mathrm{s}) \end{gathered}$$

a. We have to write a realistic problem for which these are the correct equations.

b. Finish the solution of the problem, including a pictorial representation.

A car tire is rotating at $2.5\mathrm{rad}$. After traveling $50\mathrm{m}$in $10\mathrm{s}$with the acceleration of $1.5\mathrm{m}/{\mathrm{s}}^2$.What is the car tire's initial and final velocity?

We are given with two equations, from first equation:

$$\begin{gathered} 2.5\mathrm{rad}=0\mathrm{rad}+{\mathrm{Ó\neg }}_{\mathrm{i}}(10\mathrm{s})+\left(\right(1.5\mathrm{m}/{\mathrm{s}}^2)/2(50\mathrm{m}\left)\right){(10\mathrm{s})}^2 \\ 2.5\mathrm{rad}={\mathrm{Ó\neg }}_{\mathrm{i}}(10\mathrm{s})+\left(\right(1.5\mathrm{m}/{\mathrm{s}}^2)/2(50\mathrm{m}\left)\right)100{\mathrm{s}}^2 \\ 2.5\mathrm{rad}={\mathrm{Ó\neg }}_{\mathrm{i}}(10\mathrm{s})+1.5 \\ 1\mathrm{rad}=\mathrm{Ó\neg i}(10\mathrm{s}) \\ {\mathrm{Ó\neg }}_{\mathrm{i}}=0.1\mathrm{s} \end{gathered}$$

From second equation:

$$\begin{gathered} {\mathrm{Ó\neg }}_{\mathrm{f}}={\mathrm{Ó\neg }}_{\mathrm{i}}+\left(\right(1.5\mathrm{m}/{\mathrm{s}}^2)/(50\mathrm{m}\left)\right)(10\mathrm{s}) \\ {\mathrm{Ó\neg }}_{\mathrm{f}}=0.1\mathrm{rad}/\mathrm{s}+\left(\right(1.5\mathrm{m}/{\mathrm{s}}^2)/(50\mathrm{m}\left)\right)(10\mathrm{s}) \\ {\mathrm{Ó\neg }}_{\mathrm{f}}=0.1\mathrm{rad}/\mathrm{s}+0.3 \\ {\mathrm{Ó\neg }}_{\mathrm{f}}=0.4\mathrm{rad}/\mathrm{s} \end{gathered}$$

The Pictorial representation of the given equations are as: