$1.3.24.$ At a time when the particle velocity is $10^6$ $\frac{m}{s}$, its acceleration is $10^4$ $\frac{m}{s^2}$ and is directed at an angle of $30^\circ$ to the velocity. How much will the speed increase in $10^{-2}$ seconds? At what angle will the speed direction change? What is the angular velocity of rotation of the velocity vector at this moment?
We decompose the acceleration into two components, tangential $a_\parallel $ and normal $a_\perp$:
$$ a_\parallel = a \cdot \cos 30^{\circ} = a \frac{\sqrt{3}}{2} $$
$$ a_\perp = a \cdot \sin 30^{\circ}= a \frac{1}{2} $$
During time $\Delta t$ the speed will increase by $\Delta v_\perp = a_\perp \Delta t$ and $\Delta v _\parallel = a_\parallel \Delta t$, in the transverse and longitudinal directions, respectively.
New speed values:
$$ v_\parallel = \frac{\sqrt{3}}{2}a \Delta t +v_0 $$
$$ v_\perp = \frac{1}{2}a \Delta t $$
Thus, the desired angle:
$$ \tan \alpha = \frac{v_\perp}{v_\parallel} $$
Angular velocity
$$ \omega = \frac{\alpha}{t} $$
$\text{With}(\sqrt{3}/2)\cdot10^2\text{ m/s;}$ на $5\cdot10^{-5}\text{ rad;}$ $\omega=5\cdot10^{-3}\text{ s}^{-1}.$