Question

Two trains one of length \(100 ~𝑚\) and another of length \(125~ 𝑚\), are moving in mutually opposite directions along parallel lines, meet each other, each with speed \(10 ~𝑚/𝑠\). If their acceleration are \(0.3~ 𝑚/𝑠^2\) and \(0.2 ~𝑚/𝑠^2\) respectively, then the time they take to pass each other will be

Answer 1

Given: Length of train \(𝐴, 𝑠_1=100~ 𝑚 \)
Acceleration of train \(𝑎_𝐴=0.3 ~𝑚/𝑠^2\)
Length of train \(𝐵, 𝑠_2=125 ~𝑚\)
Acceleration of train \(𝑎_𝐵=0.2 ~𝑚/𝑠^2\)
Relative velocity of train \(𝐴~ 𝑤.𝑟.𝑡.\) train\(𝐵\)
\(𝑉_{AB}=𝑉_𝐴−𝑉_𝐵=10−(−10)\)
\(𝑉_{AB}=20~ 𝑚/𝑠\).
As per the question
\(𝑎_𝐴=0.3 ~𝑚/𝑠^2\\ 𝑎_𝐵=0.2 ~𝑚/𝑠^2\)
Therefore,
Relative acceleration
\(𝑎_{AB}=0.3~m/s^2+0.2~m/s^2\)
\(𝑎_{AB}=0.5~ 𝑚/𝑠^2\)
As we know, from \(2^{nd}\) equation of motion we get,
\(𝑠=𝑢𝑡+\dfrac{1}{2}~ 𝑎𝑡^2\) …(𝑖)
As given,
\( s~=~s_1+𝑠_2=100~m+125~m=225~ 𝑚\)
By putting the values in equation (𝑖) we get,
\(225=20t+\dfrac{1}{2}×0.5×𝑡^2\)
\(0.5𝑡^2+40𝑡−450=0\)
\( 𝑡=\dfrac{−40±\sqrt{1600+4×(0.5)×450}}{1}\)
\(=−40±50\)
Therefore,
\(𝑡=10~s ~\text{(Taking} +𝑣𝑒~ \text{value}).\)