Zadanie 4 - Mechanika relatywistyczna - dylatacja czasu, droga relatywistyczna

Dane:
t_{0}=2,5 \cdot 10^{-8} s
v_{1}=0,7c
v_{2}=0,99c
c=3 \cdot 10^8 \frac{m}{s}

t^{'}_{1}=\frac{t_{0}}{\sqrt{1-\frac{v^2_1}{c^2}}}

t^{'}_{1}=\frac{t_{0}}{\sqrt{1-\frac{(0,7c)^2}{c^2}}}

t^{'}_{1}=\frac{t_{0}}{\sqrt{1-\frac{0,49c^2}{c^2}}}

t^{'}_{1}=\frac{t_{0}}{\sqrt{1-0,49}}

t^{'}_{1}=\frac{t_{0}}{\sqrt{0,51}}

t^{'}_{1}= \frac{t_{o}}{0,714142842854285}

t^{'}_{1}=1,4002800840280098031370195338695t_{o}

t^{'}_{1} \approx 1,4t_{o}

t^{'}_{1} \approx 1,4 \cdot 2,5 \cdot 10^{-8} s

t^{'}_{1} \approx 3,5 \cdot 10^{-8} s

t^{'}_{2}=\frac{t_{0}}{\sqrt{1-\frac{v^2_2}{c^2}}}

t^{'}_{2}=\frac{t_{0}}{\sqrt{1-\frac{(0,99c)^2}{c^2}}}

t^{'}_{2}=\frac{t_{0}}{\sqrt{1-\frac{0,9801c^2}{c^2}}}

t^{'}_{2}=\frac{t_{0}}{\sqrt{1-\frac{0,9801}{1}}}

t^{'}_{2}=\frac{t_{0}}{\sqrt{1-0,9801}}

t^{'}_{2}=\frac{t_{0}}{\sqrt{0,0199}}

t^{'}_{2}=50,251256281407035175879396984925t_{0}

t^{'}_{2} \approx 50,25t_{0}

t^{'}_{2} \approx 50,25 \cdot 2,5 \cdot 10^{-8} s

t^{'}_{2} \approx 125,625 s

S_{1}= v_{1} \cdot t^{'}_{1}

S_{1}= 0,7c \cdot t^{'}_{1}

S_{1}= 0,7 \cdot 3 \cdot 10^8 \frac{m}{s} \cdot 3,5 \cdot 10^{-8} s = 7,35m

S_{2}= v_{2} \cdot t^{'}_{2}

S_{2}= 0,99c \cdot t^{'}_{2}

S_{2}= 0,99 \cdot 3 \cdot 10^8 \frac{m}{s} \cdot 125,625 s

S_{2}= 373,10625m



Standardowo, jeżeli są jakieś pytania prosimy o komentarze.