Ćwiczenia 23.03.09

Zadanie 1

$$|\overrightarrow{V}|\ = \ \sqrt{{V_{1}}^{2}\ + \ {(V_{0}\ + \ sin\frac{\pi y}{L}\ )}^{2}}$$

$V_{2}\ = \ V_{0}\text{\ sin}\frac{\pi y}{L}$

$$y\ = \ \int_{0}^{t}{V_{y}\text{\ dt}}\ = \ V_{1}(t)|\begin{matrix} t \\ 0 \\ \end{matrix}\ = \ V_{1}t$$

$$x\ = \ \int_{0}^{t}{V_{x}\text{\ dt}} = \ \int_{0}^{t}{V_{0}\sin{\frac{\pi V_{1}t}{L}\ dt =}}V_{0}\frac{L}{\pi V_{1}}\int_{}^{}{\sin z\ dz = \ - \frac{V_{0}L}{\pi V_{1}}\cos{\frac{\pi V_{1}t}{L}\ |\begin{matrix} + 0 \\ 0 \\ \end{matrix} = \frac{V_{0}L}{\pi V_{1}}(1 - \cos\frac{\text{πt}V_{0}}{L})}}$$

$$x = \frac{V_{0}L}{\pi V_{1}}(1 - \cos\frac{\text{πt}}{L})$$

Zadanie 2

$$y\ = \frac{1}{2}\text{L\ }t_{0}\ V_{x}\ = \ u\frac{2L}{2L}$$

$y = \frac{L}{2}$ Vx = u 

y = L Vx = 0

0  =  a L  +  b $U\ = \ a\frac{L}{2}\ \ aL$

$U\ = \ a\frac{L}{2}\ + \ b$ $a\ = \ - \frac{2U}{L}$

V – const.

$$Vx\ = \ u\frac{2y}{L}\backslash t,\ \ \ \text{y\ } \in (0,\frac{L}{2})$$

$$Vx\ = \ - 2\frac{U}{L}\ y\ + \ 2U\backslash t,\ \ \text{y\ } \in \ (\frac{L}{2},\ L)$$

y =  V t

$$x\ = \ \int_{0}^{t_{1}}\text{Vx\ dt}\ = \ \int_{0}^{t_{2}}{U\frac{2}{L}\text{\ Vt\ dt}}\ ,\ \ \text{y\ } \in \ (0,\frac{L}{2})$$

$$x = \frac{2UV}{L}\text{\ \ }\int_{0}^{t_{1}}\text{\ t\ dt}\ = \frac{2UV}{L}\ \ \frac{{t_{1}}^{2}}{2}\ = \frac{Uy^{2}}{L}$$