Question

Short answer

The results for parts (a), (b), and (c) are$-3.1\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2$, $12.2\mathrm{s}$, and $282.4\mathrm{m}$,respectively.

Step-by-step solution

Given data

The number of revolutions is$\theta =75\mathrm{r}\mathrm{e}\mathrm{v}$.

The initial speed of the car is$v_1=95\mathrm{k}\mathrm{m}/\mathrm{h}$.

The final speed of the car is$v_2=55\mathrm{k}\mathrm{m}/\mathrm{h}$.

The diameter of the tire is$D=0.80\mathrm{m}$.

Understanding angular velocity

In this problem, first, the angular velocity of the wheel needs to be calculated by dividing the linear velocity with time. Thereafter, the equation of kinematics needs to be used to evaluate the angular acceleration.

Determine the angular acceleration

The angular acceleration can be calculated as:

$\begin{array}{rl}l\alpha & =\frac{{{\omega }_2}^2-{{\omega }_1}^2}{2\theta }\\ \alpha & =\frac{{v_2}^2-{v_1}^2}{2\theta r^2}\\ \alpha & =\frac{{v_2}^2-{v_1}^2}{2\theta {\left(\frac{D}{2}\right)}^2}\end{array}$

On plugging the values in the above relation, you get:

$\begin{array}{rl}l\alpha & =\left(\frac{{\left(55\mathrm{k}\mathrm{m}/\mathrm{h}\right)}^2-{\left(95\mathrm{k}\mathrm{m}/\mathrm{h}\right)}^2{\left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{k}\mathrm{m}/\mathrm{h}}\right)}^2}{2\left(75\mathrm{r}\mathrm{e}\mathrm{v}\times \frac{2\pi \mathrm{r}\mathrm{a}\mathrm{d}}{1\mathrm{r}\mathrm{e}\mathrm{v}}\right){\left(\frac{0.8\mathrm{m}}{2}\right)}^2}\right)\\ \alpha & =-3.1\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2\end{array}$

Thus, $\alpha =-3.1\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2$ is the required angular acceleration.

Determine the time to stop the car

The relation to obtain the required time is given by:

$t=\frac{\omega -{\omega }_0}{\alpha }$

Here, ${\omega }_0$ and $\omega$ are the initial and final angular velocities, where the final angular velocity is equal to zero.

On plugging the values in the above relation, you get:

$\begin{array}{rl}lt& =\frac{0-v_2}{\alpha r}\\ t& =-\frac{v_2}{\alpha \left(\frac{D}{2}\right)}\\ t& =-\left(\frac{\left(55\mathrm{k}\mathrm{m}/\mathrm{h}\right)\left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{k}\mathrm{m}/\mathrm{h}}\right)}{\left(-3.1\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2\right)\left(\frac{0.8\mathrm{m}}{2}\right)}\right)\\ t=12.2\mathrm{s}\end{array}$

Thus, $t=12.2\mathrm{s}$ is the required time.

Determine the total angular displacement

The relation to obtain the required time is given by:

${\omega }^2={{\omega }_0}^2+2\alpha \mathrm{\Delta }\theta$

On plugging the values in the above relation, you get:

$\begin{array}{rl}l{\left(0\right)}^2& ={\left(\frac{v_2}{\left(\frac{D}{2}\right)}\right)}^2+2\alpha \mathrm{\Delta }\theta \\ \mathrm{\Delta }\theta & =\frac{{\left(\frac{\left(55\mathrm{k}\mathrm{m}/\mathrm{h}\right)\left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{k}\mathrm{m}/\mathrm{h}}\right)}{\left(\frac{0.8\mathrm{m}}{2}\right)}\right)}^2}{2\left(-3.1\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2\right)}\\ \mathrm{\Delta }\theta & =235.2\mathrm{r}\mathrm{a}\mathrm{d}\end{array}$

Determine the total distance

The relation to obtain the distance traveled by the car is given by:

$\begin{array}{rl}l{d}_1& =r\mathrm{\Delta }\theta \\ d_1& =\left(\frac{D}{2}\right)\mathrm{\Delta }\theta \\ d_1& =\left(\frac{0.8\mathrm{m}}{2}\right)\left(235.2\mathrm{r}\mathrm{a}\mathrm{d}\right)\\ d_1& =94.08\mathrm{m}\end{array}$

The relation to obtain the distance covered by the car during 75 revolutions is given by:

$\begin{array}{rl}l{d}_2& =r\theta \\ d_2& =\left(\frac{D}{2}\right)\theta \\ d_2& =\left(\frac{0.8\mathrm{m}}{2}\right)\left(75\mathrm{r}\mathrm{e}\mathrm{v}\times \frac{2\pi \mathrm{r}\mathrm{a}\mathrm{d}}{1\mathrm{r}\mathrm{e}\mathrm{v}}\right)\\ d_2& =188.4\mathrm{m}\end{array}$

The total distance traveled by the car is calculated as:

$\begin{array}{rl}l{d}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}& =d_1+d_2\\ d_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}& =\left(94.08\mathrm{m}\right)+\left(188.4\mathrm{m}\right)\\ d_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}& =282.4\mathrm{m}\end{array}$

Thus, $d_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=282.4\mathrm{m}$ is the total distance.