User:Dionyziz/Physics

B6
$$\cos\theta = \frac{2ct}{\sqrt{b^2 + 2c^2t^2}}$$

$$\cos\theta = \frac{2ct}{\sqrt{b^2 + 4c^2t^2}}$$

Older stuff
$$u(t) = \sqrt{2g(\frac{x^2(t)}{L} - \frac{b^2}{L} - x(t) + b)}$$

B4
$$ \frac{du}{dx} \cdot \frac{dx}{dt} =- \frac{k \cdot u^2(t)}{m}$$

$$ \frac{1}{u} du = - \frac{k}{m} dx$$

$$ \int_{u_0}^{u}{\frac{1}{u} du} = - \int_{0}^{x}{\frac{k}{m} dx}$$

$$ ln[u(x)] - ln{u_0} = - \frac{k}{m} \cdot x $$

$$ u(x) = u_0 \cdot e^{- \frac{k}{m} \cdot x }$$

$$ m \cdot a(t)=-k{u(t)}^2$$

$$ a(t) = - \frac{k \cdot u^2(t)}{m}$$

$$ \frac{du}{dt} = - \frac{k \cdot u^2(t)}{m}$$

$$ - \frac{1}{u^2(t)} du = \frac{k}{m}dt$$

$$\int_{u_0}^{u}{- \frac{1}{u^2(t)} du } = \int_{0}^{t}{\frac{k}{m}dt}$$

$$ \frac{1}{u(t)} - \frac{1}{u_0} = \frac{k}{m} \cdot t$$

$$u(t)=\frac{1}{ \frac{1}{V_0} + \frac{k}{m} \cdot t} = \frac{m \cdot V_0}{m+V_0k \cdot t}$$

$$x(t)=\frac{m}{k} ln(m+V_0 \cdot k \cdot t)$$

$$\vec{a}(t) = \vec{g}\frac{2x(t)}{L} - \vec{g}$$

$$\frac{du}{dt} = \frac{g}{L} (2x - L)$$

$$\frac{du}{dx} \frac{dx}{dt} = \frac{g}{L} (2x - L)$$

$$u \frac{du}{dx} = \frac{g}{L} (2x - L)$$

$$\int_{0}^{u}{u du} = \frac{g}{L} \int_{b}^{x}{(2x - L)dx}$$

$$\left[ \frac{u^2}{2} \right]_{0}^{u(t)} = \frac{g}{L} (\int_{b}^{x}{2x} - \int_{b}^{x}{Ldx})$$

$$\frac{u^2(t)}{2} = \frac{g}{L} (\int_{b}^{x}{2x} - \int_{b}^{x}{Ldx})$$

$$\frac{u^2(t)}{2} = \frac{g}{L} (\left[ x^2 \right]_b^x - L(x(t) - b))$$

$$\frac{u^2(t)}{2} = \frac{g}{L} (x^2 - b^2- L(x(t) - b))$$

$$u(\tau) = \sqrt{\frac{2g}{L} (x^2 - b^2- L(\frac{2L}{3} - b))}$$

$$u(\tau) = \sqrt{2g (- \frac{b^2}{L} - \frac{2L}{9} + b)}$$

b
$$\frac{m}{k}(\ln(kt + \frac{m}{u_0}) + c) = x$$

$$c = -\ln\frac{m}{u_0}$$

$$x = \frac{m}{k}(\ln(kt + \frac{m}{u_0}) - \ln\frac{m}{u_0})$$

$$x = \frac{m}{k}\ln\frac{kt + \frac{m}{u_0}}{\frac{m}{u_0}}$$

$$x = \frac{m}{k}\ln(\frac{ktu_0}{m} + 1)$$

$$x\frac{k}{m} = \ln(\frac{ktu_0}{m} + 1)$$

$$e^{\frac{k}{m}x} = \frac{ktu_0}{m} + 1$$

$$m(\frac{e^{\frac{k}{m}x} - 1}{ku_0}) = t$$