Answer
$\Delta t=0.285s$
Work Step by Step
We have $v_{max}=A\omega$ and $a_{max}=A\omega^2$. Therefore, $$\frac{a_{max}}{v_{max}}=\omega=\frac{6.89}{1.25}=5.51rad/s$$
Recall that $v=-A\omega\sin\theta$ and $a=-A\omega^2\cos\theta$
So $v_{max}$ when $\sin\theta=1$, or $\theta=90^o=\frac{\pi}{2}rad$
And $a_{max}$ when $\cos\theta=1$, or $\theta=0^o=0rad$
The difference in angular displacement between these 2 points is $\Delta\theta=\frac{\pi}{2}rad$
Since we know $\omega=5.51rad/s$, we can find out how much time elapses between these 2 points: $$\Delta t=\frac{\Delta\theta}{\omega}=0.285s$$
