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

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.