91Ó°ÊÓ

1. The radius is, r = 30.0m.
2. The constant acceleration of car is, a = 0.500m/s².
3. Time is t = 15.0s.

By using kinematic equation, we can find the linear velocity v.By usingthis v, we can find the angular velocity $\mathit{Ó\neg }$and angular acceleration $\mathit{Î\pm }$, and using all these values, we can find the magnitude of net linear acceleration15 slater and theangle this net acceleration vector makes with the car’s velocity.

1. The kinematic equation is $\mathit{v}\mathbf{=}{\mathit{v}}_{\mathbf{0}}\mathbf{+}\mathit{a}\mathit{t}$
2. The angular velocity is $\mathit{Ó\neg }\mathbf{=}\frac{\mathbf{v}}{\mathbf{r}}$
3. The angular acceleration is $\mathit{Î\pm }\mathbf{=}\frac{\mathbf{Ó\neg }}{\mathbf{t}}$
4. The tangential acceleration is${\mathit{a}}_{\mathbf{t}}\mathbf{=}\mathit{Î\pm }\mathit{r}$
5. The radial acceleration is ${\mathit{a}}_{\mathbf{r}}\mathbf{=}{\mathit{Ó\neg }}^{\mathbf{2}}\mathit{r}$
6. The magnitude of acceleration is role="math" localid="1661063908108" $\mathbf{\left|}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{a}}\mathbf{\right|}\mathbf{=}\sqrt{({a_r}^2+{a_t}^2)}$
7. The angle this net acceleration vector makes with the car’s velocity is $\mathit{θ}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{a_r}{a_t}\right)$

We know that the kinematic equation is

$v=v_0+at$

But $v_0=0$, so the linear velocity v is given by

$v=at$

Substitute the all the value in the above equation.

$\begin{array}{rcl}v& =& 0.500m/s^2×15s\\ & =& 7.50m/s\end{array}$

The angular velocity $\mathrm{Ó\neg }$is given by

$\mathrm{Ó\neg }=\frac{v}{r}$

Substitute the all the value in the above equation.

$\begin{array}{rcl}\mathrm{Ó\neg }& =& \frac{7.50m/s}{30.0m}\\ & =& 0.250rad/s\end{array}$

Now, the angular acceleration is given by

$\mathrm{Î\pm }=\frac{\mathrm{Ó\neg }}{t}$

Substitute the all the value in the above equation.

$\begin{array}{rcl}\mathrm{Î\pm }& =& \frac{0.250rad/s}{15s}\\ & =& 0.017rad/s^2\end{array}$

The radial acceleration is given by

$a_r={\mathrm{Ó\neg }}^{2}r$

Substitute the all the value in the above equation.

$\begin{array}{rcl}{a}_r& =& {\left(0.250rad/s^2\right)}^2×30.0m\\ & =& 1.875m/s^2\end{array}$

The tangential acceleration is given by

$a_t=\mathrm{Î\pm }r$

Substitute the all the value in the above equation.

$\begin{array}{rcl}{a}_t& =& 0.0166rad/s^2×30.0m\\ & =& 0.50m/s^2\end{array}$

Thus, the magnitude of net linear acceleration is

$\left|\stackrel{â‡\P Ä}{\mathrm{a}}\right|=\sqrt{({{\mathrm{a}}_{\mathrm{r}}}^2+{{\mathrm{a}}_{\mathrm{t}}}^2)}$

Substitute the all the value in the above equation.

$\begin{array}{rcl}\left|\stackrel{â‡\P Ä}{\mathrm{a}}\right|& =& \sqrt{{\left(1.875\mathrm{m}/{\mathrm{s}}^2\right)}^2+{\left(0.50\mathrm{m}/{\mathrm{s}}^2\right)}^2}\\ & =& 1.94m/s^2\end{array}$

Hence the linear acceleration is, 1.94m/s².

The angle this net acceleration vector makes with the car’s velocity at this time is

$\mathrm{θ}={\tan }^{-1}\left(\frac{a_r}{a_t}\right)$

Substitute the all the value in the above equation.

$\begin{array}{rcl}\mathrm{θ}& =& {\tan }^{-1}\left(\frac{1.875m/s^2}{0.500m/s^2}\right)\\ & =& 75.1^o\end{array}$

Hence the angle is, 75.1^o.