A) [tex]+1.67 rad/s^2[/tex]
The angular acceleration of the wheel is given by
[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]
where
[tex]\omega_i = -6.00 rad/s[/tex] is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)
[tex]\omega_f = 4.00 rad/s[/tex] is the final angular velocity (anticlockwise, so with a positive sign)
[tex]\Delta t= 6.00 s - 0=6.00 s[/tex] is the time interval
Substituting into the equation, we find the angular acceleration:
[tex]\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2[/tex]
And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.
B) 3.6 s
The time interval during which the angular velocity is increasing is the time interval between the instant [tex]t_1[/tex] where the angular velocity becomes positive (so, [tex]\omega_i=0[/tex]) and the time corresponding to the final instant [tex]t_2 = 6.0 s[/tex], where [tex]\omega_f = +6.00 rad/s[/tex]. We can find this time interval by using
[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]
And solving for [tex]\Delta t[/tex] we find
[tex]\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s[/tex]
C) 2.4 s
The time interval during which the angular velocity is idecreasing is the time interval between the initial instant [tex]t_1=0[/tex] when [tex]\omega_i=-4.00 rad/s[/tex]) and the time corresponding to the instant in which the velovity becomes positive [tex]t_2[/tex], when [tex]\omega_f = 0 rad/s[/tex]. We can find this time interval by using
[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]
And solving for [tex]\Delta t[/tex] we find
[tex]\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s[/tex]
D) 5.6 rad
The angular displacement of the wheel is given by the equation
[tex]\omega_f^2 - \omega_i^2 = 2 \alpha \theta[/tex]
where we have
[tex]\omega_i = -6.00 rad/s[/tex] is the initial angular velocity of the wheel
[tex]\omega_f = 4.00 rad/s[/tex] is the final angular velocity
[tex]\alpha=+1.67 rad/s^2[/tex] is the angular acceleration
Solving for [tex]\theta[/tex],
[tex]\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad[/tex]
