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?

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    <h3> Statement </h3>
    <p>
        $1.3.24.$ 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?
    </p>

    <h3>Solution</h3>
    <p>
        <p>We decompose the acceleration into two components, tangential $a_\parallel $ and normal $a_\perp$:</p>

$$a_\parallel = a \cdot \cos 30^{\circ} = a \frac{\sqrt{3}}{2}$$

$$a_\perp = a \cdot \sin 30^{\circ}= a \frac{1}{2}$$

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

$$\omega = \frac{\alpha}{t}$$

    <h4>Answer</h4>
    <p>
        $\text{With}(\sqrt{3}/2)\cdot10^2\text{ m/s;}$ на $5\cdot10^{-5}\text{ rad;}$ $\omega =5\cdot10^{-3}\text{ s}^{-1}.$
    </p>



    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Aliaksandr Melnichenka<br>
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