Question

A train starts from station A with uniform acceleration \[{a_1}\], for some distance and then goes with uniform retardation \[{a_2}\] for some more distance to come to rest at station B. The distance between station A and B is \[4\,{\text{km}}\] and the train takes \[\dfrac{1}{{15}}\,{\text{h}}\] to complete this journey. If the accelerations are in km per minute, find the value of \[\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}}\].

Answer 1

Hint: The kinematic expression relating displacement \[s\], initial velocity \[u\], acceleration \[a\] and time \[t\] is \[s = ut + \dfrac{1}{2}a{t^2}\] …… (1)
The kinematic expression relating final velocity \[v\], initial velocity \[u\], acceleration \[a\] and time \[t\] is \[v = u + at\] …… (2)

Complete step by step answer:
The train starts moving from station A with uniform acceleration \[{a_1}\]. Then after travelling some distance, it starts moving with the uniform retardation \[{a_2}\] upto station B. The distance between the station A and B is \[4\,{\text{km}}\] and it takes \[\dfrac{1}{{15}}\,{\text{h}}\] for the train to cover this distance.
Consider the travel of the train from station A and cover some distance\[{s_1}\] with the velocity \[{v_1}\] in time \[{t_1}\].
The initial velocity \[{u_1}\] of the train at station A is zero.
\[{u_1} = 0\,{\text{m/s}}\]
Rewrite equation (2) for the final velocity of the train from station A.
\[{v_1} = {u_1} + {a_1}{t_1}\]
Substitute \[0\,{\text{m/s}}\] for \[{u_1}\] in the above equation and rearrange it for \[{a_1}\].
\[{v_1} = \left( {0\,{\text{m/s}}} \right) + {a_1}{t_1}\]
\[ \Rightarrow {a_1} = \dfrac{{{v_1}}}{{{t_1}}}\]
Rewrite equation (2) for the displacement \[{s_1}\] of the train from station A in its first travel.
\[{s_1} = {u_1}{t_1} + \dfrac{1}{2}{a_1}t_1^2\]
Substitute \[0\,{\text{m/s}}\] for \[{u_1}\]and \[\dfrac{{{v_1}}}{{{t_1}}}\] for \[{a_1}\] in the above equation.
\[{s_1} = \left( {0\,{\text{m/s}}} \right){t_1} + \dfrac{1}{2}\dfrac{{{v_1}}}{{{t_1}}}t_1^2\]
\[ \Rightarrow {s_1} = \dfrac{1}{2}{v_1}{t_1}\]
This is the displacement of the train in its first travel.
After the displacement \[{s_1}\] of the train, the train starts moving with the uniform retardation \[{a_2}\] to cover the remaining distance \[{s_2}\] upto station B with velocity \[{v_2}\] in time \[{t_2}\].
The final velocity \[{v_1}\] of the train in its first travel is the initial velocity \[{u_2}\] of the train in its second travel.
\[{u_2} = {v_1}\]
The train stops at station B. Hence, the final velocity of the train at station B is zero.
Rewrite equation (2) for the final velocity of the train towards station B.
\[{v_2} = {u_2} + {a_2}{t_2}\]
Substitute \[0\,{\text{m/s}}\] for \[{v_2}\], \[{v_1}\] for \[{u_2}\] in the above equation and rearrange it for \[{a_2}\].
\[0\,{\text{m/s}} = {v_1} - {a_2}{t_2}\]
\[ \Rightarrow {a_2} = \dfrac{{{v_1}}}{{{t_2}}}\]
The negative sign of the acceleration indicates that the acceleration \[{a_2}\] is decreasing.
Rewrite equation (2) for the displacement \[{s_2}\] of the train upto station B in its second travel.
\[{s_2} = {u_2}{t_2} - \dfrac{1}{2}{a_2}t_2^2\]
Substitute \[{v_1}\] for \[{u_2}\]and \[\dfrac{{{v_1}}}{{{t_2}}}\] for \[{a_2}\] in the above equation.
\[{s_2} = {v_1}{t_2} - \dfrac{1}{2}\dfrac{{{v_1}}}{{{t_2}}}t_2^2\]
\[ \Rightarrow {s_2} = {v_1}{t_2} - \dfrac{1}{2}{v_1}{t_2}\]
\[ \Rightarrow {s_2} = \dfrac{1}{2}{v_1}{t_2}\]
This is the displacement of the train in its second travel.
The total displacement \[s\] of the train is the sum of the displacements \[{s_1}\] and \[{s_2}\].
\[s = {s_1} + {s_2}\]
Substitute \[\dfrac{1}{2}{v_1}{t_1}\] for \[{s_1}\] and \[\dfrac{1}{2}{v_1}{t_2}\] for \[{s_2}\] in the above equation.
\[s = \dfrac{1}{2}{v_1}{t_1} + \dfrac{1}{2}{v_1}{t_2}\]
\[ \Rightarrow s = \dfrac{1}{2}{v_1}\left( {{t_1} + {t_2}} \right)\]
Rearrange the above equation for \[{v_1}\].
\[{v_1} = \dfrac{{2s}}{{{t_1} + {t_2}}}\]
Here, \[s\] is the total displacement of the train and \[{t_1} + {t_2}\] is the total travel time of the train.
Substitute \[4\,{\text{km}}\] for \[s\] and \[\dfrac{1}{{15}}\,{\text{h}}\] for \[{t_1} + {t_2}\] in the above equation.
\[{v_1} = \dfrac{{2\left( {4\,{\text{km}}} \right)}}{{\dfrac{1}{{15}}\,{\text{h}}}}\]
\[ \Rightarrow {v_1} = \dfrac{{2\left( {4\,{\text{km}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{m}}}}{{1\,{\text{km}}}}} \right)}}{{\left( {\dfrac{1}{{15}}\,{\text{h}}} \right)\left( {\dfrac{{3600\,{\text{s}}}}{{1\,{\text{h}}}}} \right)}}\]
\[ \Rightarrow {v_1} = 33.33\,{\text{m/s}}\]
Hence, the final velocity of the train after first travel is \[33.33\,{\text{m/s}}\].
Now, calculate the value of \[\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}}\].
Substitute \[\dfrac{{{v_1}}}{{{t_1}}}\] for \[{a_1}\] and \[\dfrac{{{v_1}}}{{{t_2}}}\] for \[{a_2}\] in \[\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}}\].
\[\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} = \dfrac{{{t_1}}}{{{v_1}}} + \dfrac{{{t_2}}}{{{v_1}}}\]
\[ \Rightarrow \dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} = \dfrac{{{t_1} + {t_2}}}{{{v_1}}}\]
Substitute \[33.33\,{\text{m/s}}\] for \[{v_1}\] and \[\dfrac{1}{{15}}\,{\text{h}}\] for \[{t_1} + {t_2}\]in the above equation.
\[\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} = \dfrac{{\dfrac{1}{{15}}\,{\text{h}}}}{{33.33\,{\text{m/s}}}}\]
\[ \Rightarrow \dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} = \dfrac{{\left( {\dfrac{1}{{15}}\,{\text{h}}} \right)\left( {\dfrac{{3600\,{\text{s}}}}{{1\,{\text{h}}}}} \right)}}{{33.33\,{\text{m/s}}}}\]
\[ \Rightarrow \dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} = 7.2\]
Hence, the value of \[\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}}\] is 7.2.

Note: Take the value of acceleration for the second travel negative as it is retardation.
There are two case in given question:
1. when train moves with uniform acceleration
2. When the train moves with uniform retardation.