Answer:
76.4035 m
Explanation:
r = Radius = 0.32 m
[tex]\omega_f[/tex] = Final angular velocity = 0
[tex]\omega_i[/tex] = Initial angular velocity = 92 km/h
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation
Angular speed is given by
[tex]\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{\dfrac{92}{3.6}}{0.32}\\\Rightarrow \omega=79.861\ rad/s[/tex]
The angular speed of the tires about their axles is 79.861 rad/s.
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-79.861^2}{2\times 2\pi \times 38}\\\Rightarrow \alpha=-13.355\ rad/s^2[/tex]
The magnitude of acceleration is 13.355 m/s²
Distance is given by
[tex]d=\theta r\\\Rightarrow d=38\times 2\pi\times 0.32\\\Rightarrow d=76.4035\ m[/tex]
The distance moved while slowing down is 76.4035 m
