91Ó°ÊÓ

The number of revolutions is\(\theta = 75\;{\rm{rev}}\).

The initial speed of the car is\({v_1} = 95\;{\rm{km/h}}\).

The final speed of the car is\({v_2} = 55\;{\rm{km/h}}\).

The diameter of the tire is\(D = 0.80\;{\rm{m}}\).

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.

The angular acceleration can be calculated as:

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

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

\(\begin{aligned}{l}\alpha &= \left( {\frac{{{{\left( {55\;{\rm{km/h}}} \right)}^2} - {{\left( {95\;{\rm{km/h}}} \right)}^2}{{\left( {\frac{{1\;{\rm{m/s}}}}{{3.6\;{\rm{km/h}}}}} \right)}^2}}}{{2\left( {75\;{\rm{rev}} \times \frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}}} \right){{\left( {\frac{{0.8\;{\rm{m}}}}{2}} \right)}^2}}}} \right)\\\alpha &= - 3.1\;{\rm{rad/}}{{\rm{s}}^2}\end{aligned}\)

Thus, \(\alpha = - 3.1\;{\rm{rad/}}{{\rm{s}}^2}\) is the required angular acceleration.

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:

\(\begin{aligned}{l}t &= \frac{{0 - {v_2}}}{{\alpha r}}\\t &= - \frac{{{v_2}}}{{\alpha \left( {\frac{D}{2}} \right)}}\\t &= - \left( {\frac{{\left( {55\;{\rm{km/h}}} \right)\left( {\frac{{1\;{\rm{m/s}}}}{{3.6\;{\rm{km/h}}}}} \right)}}{{\left( { - 3.1\;{\rm{rad/}}{{\rm{s}}^2}} \right)\left( {\frac{{0.8\;{\rm{m}}}}{2}} \right)}}} \right)\\t = 12.2\,{\rm{s}}\end{aligned}\)

Thus, \(t = 12.2\,{\rm{s}}\) is the required time.

The relation to obtain the required time is given by:

\({\omega ^2} = {\omega _0}^2 + 2\alpha \Delta \theta \)

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

\(\begin{aligned}{l}{\left( 0 \right)^2} &= {\left( {\frac{{{v_2}}}{{\left( {\frac{D}{2}} \right)}}} \right)^2} + 2\alpha \Delta \theta \\\Delta \theta &= \frac{{{{\left( {\frac{{\left( {55\;{\rm{km/h}}} \right)\left( {\frac{{1\;{\rm{m/s}}}}{{3.6\;{\rm{km/h}}}}} \right)}}{{\left( {\frac{{0.8\;{\rm{m}}}}{2}} \right)}}} \right)}^2}}}{{2\left( { - 3.1\;{\rm{rad/}}{{\rm{s}}^2}} \right)}}\\\Delta \theta &= 235.2\;{\rm{rad}}\end{aligned}\)

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

\(\begin{aligned}{l}{d_1} &= r\Delta \theta \\{d_1} &= \left( {\frac{D}{2}} \right)\Delta \theta \\{d_1} &= \left( {\frac{{0.8\;{\rm{m}}}}{2}} \right)\left( {235.2\;{\rm{rad}}} \right)\\{d_1} &= 94.08\;{\rm{m}}\end{aligned}\)

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

\(\begin{aligned}{l}{d_2} &= r\theta \\{d_2} &= \left( {\frac{D}{2}} \right)\theta \\{d_2} &= \left( {\frac{{0.8\;{\rm{m}}}}{2}} \right)\left( {75\;{\rm{rev}} \times \frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}}} \right)\\{d_2} &= 188.4\;{\rm{m}}\end{aligned}\)

The total distance traveled by the car is calculated as:

\(\begin{aligned}{l}{d_{{\rm{total}}}} &= {d_1} + {d_2}\\{d_{{\rm{total}}}} &= \left( {94.08\;{\rm{m}}} \right) + \left( {188.4\;{\rm{m}}} \right)\\{d_{{\rm{total}}}} &= 282.4\;{\rm{m}}\end{aligned}\)

Thus, \({d_{{\rm{total}}}} = 282.4\;{\rm{m}}\) is the total distance.