Question

Two trains, each of length \[100{\text{ }}m\], moving in opposite directions along parallel lines, meet each other with speeds of \[50\;km{h^{ - 1\;}}\] and \[40\;km{h^{ - 1}}\]. If their accelerations are \[30\;cm{s^{ - 2}}\;\] and \[20\;cm{s^{ - 2}}\], respectively, find the time they will take to pass each other.

Answer 1

Hint: To solve this question, we need to find the relative velocity, relative acceleration, and relative displacement of one train with respect to the other, and then we need to substitute these values in the second equation of motion with constant acceleration. From that, we can find the time they will take to pass each other.

Complete step by step solution:
Let the two trains be named as train A and train B.
Given the length of each train is \[100{\text{ }}m\]. Therefore the relative displacement of the train is given as
\[s = 100 + 100 = 200m\]
The second thing which is needed to be found is the initial velocity\[u\]of one train relative to the other train.
\[u = {u_1} + {u_2}\].\[{u_1}\]is the initial velocity if train A and\[{u_2}\]
\[{u_1} = 50\;km{h^{ - 1\;}}\]and\[{u_2} = 40\;km{h^{ - 1}}\]
Substituting, we get
\[u = 50km{h^{ - 1\;}} + 40km{h^{ - 1\;}}\]
\[u = 90km{h^{ - 1\;}}\]
If we convert\[km{h^{ - 1\;}}\]to\[m{s^{ - 1}}\]
\[90 \times \dfrac{{1000}}{{3600}}m/s\]
\[u = 25m/s\]
The next thing is to find the relative acceleration
Relative acceleration, \[a = {a_1} + {a_2}\]. Here\[{a_1}\] is the acceleration of train A and \[{a_2}\] is the acceleration of train B.
Given that \[{a_1} = 30\;cm{s^{ - 2}}\;\] and \[{a_2} = 20\;cm{s^{ - 2}}\;\]
Therefore the relative acceleration is,
\[a = 30\;cm{s^{ - 2}} + 20\;cm{s^{ - 2}} = 50cm{s^{ - 2}}\]
Now converting centimeter to the meter we get,
\[a = 0.5m/{s^2}\]
Now we need to put all these values in the equation of motion with constant acceleration we get,
\[s = ut + \dfrac{1}{2}a{t^2}\]
\[200 = 25t + \dfrac{1}{2}0.5{t^2}\]
Rearranging the above equation to find the time t, we get,
\[t = \dfrac{{ - 50 \pm \sqrt {2500 + 4 \times 400 \times 0.5} }}{{2 \times 0.5}}\]
\[t = - 50 \pm \sqrt {3300} \]
Time cannot be negative; we shall consider only the positive value.
\[t = - 50 + 57.44\]
\[t = 7.44\sec \]
Therefore the time they will take to pass each other \[t = 7.44\sec \]


Note:
The second equation of motion can be rewritten, for the case of free fall as \[h{\text{ }} = {\text{ }}ut{\text{ }} + {\text{ }}\dfrac{1}{2}{\text{ }}g{t^2}\] where \[h\] is the height from which the object falls on the ground from rest or the maximum height to which that object reaches when thrown upwards and\[g\]is the acceleration due to gravity.