Úprava formátování 1. a 2. příkladu z FYI

KFY FYI1/Priklad01.md

@@ -34,20 +34,20 @@ $\displaystyle s = \frac{1}{2}\cancel{a_{t}} \cdot \frac{(v_{1} - v_{2})^T}{a_{t
 
 $\displaystyle T = \frac{v_{1} - v_{0}}{\frac{v_{1}^2 - v_{0}^2}{2s}} = \frac{v_{1} - v_{0}}{v_{1}^2 - v_{0}^2} \cdot 2s = \frac{\cancel{v_{1} - v_{0}}}{\cancel{(v_{1} - v_{0})}(v_{1} + v_{0})} \cdot 2s = \frac{2s}{v_{1} + v_{0}}$
 
-$a_{0} = \sqrt{ \left(\frac{v_{1}^2 - v_{0}^2}{2s}\right)^2 + \left(\frac{v_{0}^2}{R}\right)^2 }$
+$\displaystyle a_{0} = \sqrt{ \left(\frac{v_{1}^2 - v_{0}^2}{2s}\right)^2 + \left(\frac{v_{0}^2}{R}\right)^2 }$
 
-$a_{1} = \sqrt{ \left(\frac{v_{1}^2 - v_{0}^2}{2s}\right)^2 + \left(\frac{v_{1}^2}{R}\right)^2 }$
+$\displaystyle a_{1} = \sqrt{ \left(\frac{v_{1}^2 - v_{0}^2}{2s}\right)^2 + \left(\frac{v_{1}^2}{R}\right)^2 }$
 
 - $v_{0} = 54 \text{ km/h} = 15 \text{ m/s}$
 - $v_{1} = 18 \text{ km/h} = 5 \text{ m/s}$
 
-$a_{t} = \frac{5^2 - 15^2}{2 \cdot 800} \text{ m}\cdot\text{s}^{-2} = \frac{25 - 225}{1600} \text{ m}\cdot\text{s}^{-2} = -\frac{200}{1600} \text{ m}\cdot\text{s}^{-2} = -0.125 \text{ m}\cdot\text{s}^{-2}$
+$\displaystyle a_{t} = \frac{5^2 - 15^2}{2 \cdot 800} \text{ m}\cdot\text{s}^{-2} = \frac{25 - 225}{1600} \text{ m}\cdot\text{s}^{-2} = -\frac{200}{1600} \text{ m}\cdot\text{s}^{-2} = -0.125 \text{ m}\cdot\text{s}^{-2}$
 - mínus, takže vektor míří opačným směrem
 
 ### Výsledek
 
-$T = \frac{800}{5 + 15}\cdot 2s = \frac{1600}{20}s = 80s$
+$\displaystyle T = \frac{800}{5 + 15}\cdot 2s = \frac{1600}{20}s = 80s$
 
-$a_{0} = \sqrt{ (-0.125)^2 + \left(\frac{15^2}{800}\right)^2 } \text{ m}\cdot\text{s}^{-2} = 0.308 \text{ m}\cdot\text{s}^{-2}$
+$\displaystyle a_{0} = \sqrt{ (-0.125)^2 + \left(\frac{15^2}{800}\right)^2 } \text{ m}\cdot\text{s}^{-2} = 0.308 \text{ m}\cdot\text{s}^{-2}$
 
-$a_{1} = \sqrt{ (-0.125)^2 + \left(\frac{5^2}{800}\right)^2 } \text{ m}\cdot\text{s}^{-2} = 0.129 \text{ m}\cdot\text{s}^{-2}$
+$\displaystyle a_{1} = \sqrt{ (-0.125)^2 + \left(\frac{5^2}{800}\right)^2 } \text{ m}\cdot\text{s}^{-2} = 0.129 \text{ m}\cdot\text{s}^{-2}$

KFY FYI1/Priklad02.md

@@ -26,7 +26,9 @@ Podél rovnoměrně se otáčející tyče se od jejího upevnění rovnoměrně
 ### Výpočet
 
 $\displaystyle r = \sqrt{ [v_{0} \cos(\omega t) - v_{0}\omega t \sin(\omega t)]^2 + [v_{0} \sin(\omega t) + v_{0}\omega t \cos(\omega t)]^2 } = v_{0} \cdot \sqrt{ 1 + (\omega t)^2 }$
+
 $\displaystyle a_x = \frac{dv_{x}}{dt} = \frac{d}{dt}[v_{0} \cos(\omega t) - v_{0}\omega t \sin(\omega t)] = \dots = -2 \cdot v_{0} \cdot \omega \sin(\omega t) - v_{0} \cdot \omega^2 t \cos(\omega t)$
+
 $\displaystyle a_{y} = \frac{dv_{y}}{dt} = \frac{d}{dt}[v_{0} \sin(\omega t) + v_{0}\omega t \cos(\omega t)] = \dots = 2 \cdot v_{0} \cdot \omega \cos(\omega t) - v_{0} \cdot \omega^2 t \sin(\omega t)$
 
 ### Výsledek
@@ -34,6 +36,7 @@ $\displaystyle a_{y} = \frac{dv_{y}}{dt} = \frac{d}{dt}[v_{0} \sin(\omega t) + v
 $\displaystyle a = \sqrt{ a_{t}^2 + a_{n}^2 } = \sqrt{ 4 v_{0}^2 \cdot \omega^2 + (v_{0} \cdot \omega^2 t)^2 } = v_{0} \cdot \omega \cdot \sqrt{ 4 + (\omega t)^2 }$
 
 $\displaystyle a_{t} = \frac{dv}{dt} = \frac{d}{dt}[ v_{0} \cdot \sqrt{ 1 + (\omega t)^2 } ] = v_{0} \cdot \frac{1}{\cancel{2}}[1 + (\omega t)^2]^\frac{-1}{2} \cdot \cancel{2}(\omega t) \cdot \omega = \frac{v_{0} \cdot \omega^2 \cdot t}{\sqrt{ 1+(\omega t)^2 }}$
+
 $\displaystyle a_{n} = \frac{v^2}{R} \quad$ R neznáme, ale známe $\displaystyle a = \sqrt{ a^2_{t} + a^2_{n} }$
 
 $\displaystyle a_{n} = \sqrt{ a^2 - a^2_{t} } = \sqrt{ v_{0}^2 \cdot \omega^2 \cdot [4 + (\omega t)^2] - \frac{v_{0}^2 \cdot \omega^4 \cdot t^2}{1 + (\omega t)^2} } = \dots = \frac{v_{0} \cdot \omega \cdot [2 + (\omega t)^2]}{\sqrt{ 1 + (\omega t)^2 }}$