To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
[tex]\omega = \frac{v}{R}[/tex]
Where,
[tex]\omega =[/tex]Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
[tex]\alpha = \frac{\omega}{t}[/tex]
Where
[tex]\alpha =[/tex]Angular acceleration
[tex]\omega =[/tex] Angular velocity
t = Time
Our values are
[tex]v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})[/tex]
[tex]v = 16.67m/s[/tex]
[tex]r = 0.25m[/tex]
[tex]t=6s[/tex]
Replacing at the previous equation we have that the angular velocity is
[tex]\omega = \frac{v}{R}[/tex]
[tex]\omega = \frac{ 16.67}{0.25}[/tex]
[tex]\omega = 66.67rad/s[/tex]
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
[tex]\alpha = \frac{\omega}{t}[/tex]
[tex]\alpha = \frac{66.67}{6}[/tex]
[tex]\alpha = 11.11rad/s^2[/tex]
Therefore the angular acceleration of a point on the outer edge of the tires is [tex]11.11rad/s^2[/tex]
