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**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.

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