Question

Prove.

Answer 1

\(\large \color{black}{ \normalsize \text{Two trains a and b from point A and B }\hspace{.33em}\\~\\
\normalsize \text{with a speed of }~s_1 ~\normalsize \text{and }~ s_2 ~\normalsize \text{respectively. }\hspace{.33em}\\~\\
\normalsize \text{starts moving in opposite direction in a straight line. }\hspace{.33em}\\~\\
\normalsize \text{After crossing each other, train a and b }\hspace{.33em}\\~\\
\normalsize \text{took time } ~t_1~ \normalsize \text{and } ~t_2 ~\normalsize \text{respectively to reach destination B and A. } \hspace{.33em}\\~\\
\normalsize \text{Prove that } \dfrac{s_1}{s_2}=\sqrt{\dfrac{t_2}{t_1}}\hspace{.33em}\\~\\
}\)

Answer 2

\(\large \color{black}{\begin{align} \normalsize \text{the time they meet together is }\hspace{.33em}\\~\\
t_m=\dfrac{d}{s_1+s_2}
\end{align}}\)

Answer 3

\(\large \color{black}{\begin{align} \normalsize \text{also }\hspace{.33em}\\~\\
t_m=\dfrac{x}{s_1}\\~\\
t_m=\dfrac{d-x}{s_2}\\~\\
\end{align}}\)

Answer 4

t1 = d/s1 - d/(s1+s2)
t2 = d/s2 - d/(s1+s2)

t1/t2 = (1/s1-1/(s1+s2))/(1/s2-1/(s1+s2))
t1/t2 = (s2/s1)/(s1/s2)
t1/t2 = s2^2/s1^2

Answer 5

thanks!

Answer 6

\(\large \color{black}{\begin{align} t_1 &= \dfrac{d}{s_1} - \dfrac{d}{(s_1+s_2)}\\~\\
t_2& = \dfrac{d}{s_2} - \dfrac{d}{(s1+s2)}\\~\\

\dfrac{t_1}{\frac{1}{s_1} - \frac{1}{(s_1+s_2)}}& = \dfrac{t_2}{\frac{1}{s_2} - \frac{1}{(s_1+s_2)}}\\~\\
t_1s_1\dfrac{(s_1+s_2)}{(s_2)} &= t_2s_2\dfrac{(s_1+s_2)}{s_1}\\~\\
\dfrac{t_1s_1}{s_2} &= \dfrac{t_2s_2}{s_1}\\~\\
\dfrac{s_1^2}{s_2^2} &= \dfrac{t_2}{t_1}\\~\\
\it \text{orinally posted by ganeshie8}
\end{align}}\)