Answer:
[tex]t - t_1 = \frac{-\omega_o + \sqrt{\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)}}{4\alpha}[/tex]
Explanation:
After time "t" the angular position of A is given as
[tex]\theta_a = \theta_o + \omega_o t + \frac{1}{2}\alpha t^2[/tex]
now we know that B start motion after time t1
so its angular position is also same as that of position of A after same time "t"
so we have
[tex]\theta_b = \theta_o + \frac{\omega_o}{2} (t - t_1) + \frac{1}{2}(2\alpha) (t - t_1)^2[/tex]
now since both positions are same
[tex]\theta_a = \theta_b[/tex]
[tex]\omega_o t + \frac{1}{2}\alpha t^2 = \frac{\omega_o}{2}(t - t_1} + \alpha(t - t_1)^2[/tex]
[tex]2\omega_o t + \alpha t^2 = \omega_o(t - t_1) + 2\alpha(t - t_1)^2[/tex]
[tex]t - t_1 = \frac{-\omega_o + \sqrt{\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)}}{4\alpha}[/tex]
