Explanation:
Given that,
Initial linear speed of the motorcycle, u = 7.1 m/s
Final linear speed of the motorcycle, v = 1.9 m/s
Radius of the wheel, r = 0.65 m
Time, t = 6 s
(a) Let a is the linear acceleration. It can be calculated as :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{1.9-7.1}{6}[/tex]
[tex]a=-0.86\ m/s^2[/tex]
If [tex]\alpha[/tex] is the angular acceleration. The relation is given by :
[tex]\alpha =\dfrac{a}{r}[/tex]
[tex]\alpha =\dfrac{-0.86}{0.65}[/tex]
[tex]\alpha =-1.32\ rad/s^2[/tex]
(b) Let d is the linear displacement.
[tex]d=\dfrac{u+v}{2}\times t[/tex]
Let [tex]\theta[/tex] is the angular displacement
[tex]\theta=\dfrac{d}{r}=\dfrac{\dfrac{u+v}{2}\times t}{r}[/tex]
[tex]\theta=\dfrac{\dfrac{7.1+1.9}{2}\times 6}{0.65}[/tex]
[tex]\theta=41.53\ rad[/tex]
Hence, this is the required solution.
