Given,
The initial velocity of the car, u=0 m/s
The final linear velocity of the car, v=28.7 m/s
The time duration, t=8.7 s
The diameter of the tires, d=50.1 cm=0.501 m
The radius of the tires is,
[tex]\begin{gathered} r=\frac{d}{2} \\ =\frac{0.501}{2} \\ =0.25\text{ m} \end{gathered}[/tex]From the equation of motion,
[tex]v=u+at[/tex]Where a is the acceleration of the car.
On substituting the known values,
[tex]\begin{gathered} 28.7=0+a\times8.7 \\ \Rightarrow a=\frac{28.7}{8.7} \\ a=3.3\text{ m/s}^2 \end{gathered}[/tex]The angular acceleration of the tires is given by,
[tex]\alpha=\frac{a}{r}[/tex]On substituting the known values,
[tex]\begin{gathered} \alpha=\frac{3.3}{0.25} \\ =13.2\text{ rad/s}^2 \end{gathered}[/tex]As the initial linear velocity of the car was zero, the initial angular velocity, ω₀ is also zero.
From the equation of motion,
[tex]\theta=\omega_0t+\frac{1}{2}\alpha t^2[/tex]Where θ is the angular displacement of the tire.
On substituting the known values,
[tex]\begin{gathered} \theta=0+\frac{1}{2}\times13.2\times8.7^2 \\ =499.55\text{ rad} \end{gathered}[/tex]To complete one revolution, the tire has to rotate through 2π radians.
Thus the total number of revolutions made by the tires is,
[tex]\begin{gathered} N=\frac{\theta}{2\pi} \\ =\frac{499.55}{2\pi} \\ =79.51\text{ } \end{gathered}[/tex]Therefore the total number of revolutions made by the tires is 79.51 which is approximately equal to 79.3202 in the option.
