Answer:
The angular displacement of each wheel is 269.92 rad
Explanation:
Given:
Angular acceleration [tex]\alpha = 6.10 \frac{rad}{s^{2} }[/tex]
Time to pass cyclist [tex]t = 4.36[/tex] s
Angular velocity [tex]\omega _{f} = 75.2 \frac{rad}{s}[/tex]
According to the equation of kinematics,
[tex]\omega _{f} = \omega _{i} + \alpha t[/tex]
[tex]\omega _{i} = \omega _{f} - \alpha t[/tex]
[tex]\omega _{i} = 75.2 - 6.10 \times 4.36[/tex]
[tex]\omega _{f} = 48.60[/tex] [tex]\frac{rad}{s}[/tex]
For finding angular displacement,
[tex]\omega _{f} ^{2} - \omega _{i} ^{2} = 2 \alpha \theta[/tex]
Where [tex]\theta =[/tex] angular displacement,
[tex]\theta = \frac{\omega _{f}^{2} - \omega _{i} ^{2} }{2\alpha }[/tex]
[tex]\theta = \frac{5655.04 - 2361.96 }{2\times 6.10 }[/tex]
[tex]\theta = 269.92[/tex] rad
Therefore, the angular displacement of each wheel is 269.92 rad
