In accordance with the schedule, a train is to travel the distance between $A$ and $B$, equal to $20$ km, at a constant speed. It traveled half-way with the specified speed and stopped for three minutes. To arrive at point $B$ on time, it had to increase its speed by $10$ kmph for the rest of the way. Next time the train stopped half-way for five minutes. At what speed must it travel the remaining half of the distance to arrive at point $B$ in accordance with the schedule?
Given: $\;$ Distance $d\left(AB\right) = 20$ km
Let the original speed of train $= s$ kmph
Time taken by the train to reach point $B$ $= t_B = \dfrac{20}{s}$ hr $\;\;\; \cdots \; (1)$
Let half-way point be $= A'$
Then, distance $d\left(AA'\right) = 10$ km
Time taken by the train to cover $d\left(AA'\right) = t_1 = \dfrac{10}{s}$ hr $\;\;\; \cdots \; (2)$
Train stops for $t_{stop1} = 3$ min $= \dfrac{3}{60} = \dfrac{1}{20}$ hr $\;\;\; \cdots \; (3)$
New speed of train $= s_1 = s + 10$ kmph
Distance $d\left(A'B\right) = 10$ km is covered with speed $= s_1 = \left(s + 10\right)$ kmph
Time taken to cover distance $d\left(A'B\right) = t_2 = \left(\dfrac{10}{s + 10}\right)$ hr $\;\;\; \cdots \; (4)$
Time taken by the train to reach $B$ $= t_B = t_1 + t_{stop1} + t_2$
$\therefore \;$ We have from equations $(2)$, $(3)$ and $(4)$
$t_B = \dfrac{10}{s} + \dfrac{1}{20} + \dfrac{10}{s + 10}$ hr $\;\;\; \cdots \; (5)$
$\therefore \;$ We have from equations $(1)$ and $(5)$
$\dfrac{20}{s} = \dfrac{10}{s} + \dfrac{1}{20} + \dfrac{10}{s + 10}$
i.e. $\;$ $\dfrac{10}{s} - \dfrac{10}{s + 10} = \dfrac{1}{20}$
i.e. $\;$ $10 \left[\dfrac{1}{s} - \dfrac{1}{s + 10}\right] = \dfrac{1}{20}$
i.e. $\;$ $10 \left[\dfrac{s + 10 - s}{s \left(s + 10\right)}\right] = \dfrac{1}{20}$
i.e. $\;$ $\dfrac{100}{s^2 + 10s} = \dfrac{1}{20}$
i.e. $\;$ $s^2 + 10s - 2000 = 0$
i.e. $\;$ $\left(s + 50\right) \left(s - 40\right) = 0$
i.e. $\;$ $s = -50$ $\;$ or $\;$ $s = 40$
Since the speed of the train cannot be negative
$s = -50$ is not an acceptable solution
$\therefore \;$ Original speed of train $= s = 40$ kmph
$\therefore \;$ Time taken to reach $B$ $= t_B = \dfrac{20}{40} = \dfrac{1}{2}$ hr $\;\;\; \cdots \; (1a)$ [in view of equation $(1)$]
and time taken to cover distance $d\left(AA'\right) = t_1 = \dfrac{10}{40} = \dfrac{1}{4}$ hr $\;\;\; \cdots \; (2a)$ [in view of equation $(2)$]
Now, the train stops for time $= t_{stop2} = 5$ min $= \dfrac{5}{60} = \dfrac{1}{12}$ hr $\;\;\; \cdots \; (6)$
Let the new speed of train be $= s_2$ kmph
Since $A'$ is the halfway point, distance $d\left(A'B\right) = 10$ km
Now, the train covers distance $d\left(A'B\right)$ with speed $s_2$
Time taken by the train to cover $d \left(A'B\right) = t_3 = \dfrac{10}{s_2}$ hr $\;\;\; \cdots \; (7)$
$\therefore \;$ Time taken by the train to reach point $B$ $= t_B = t_1 + t_{stop2} + t_3$
$\therefore \;$ In view of equations $(2a)$, $(6)$ and $(7)$ we have
$t_B = \dfrac{1}{4} + \dfrac{1}{12} + \dfrac{10}{s_2}$ hr $\;\;\; \cdots \; (8)$
Since the train is to arrive at $B$ in accordance with the schedule
$\therefore \;$ we have from equations $(1a)$ and $(8)$,
$\dfrac{1}{4} + \dfrac{1}{12} + \dfrac{10}{s_2} = \dfrac{1}{2}$
i.e. $\;$ $\dfrac{10}{s_2} = \dfrac{1}{2} - \dfrac{1}{4} - \dfrac{1}{12} = \dfrac{1}{6}$
i.e. $\;$ $s_2 = 60$
$\therefore \;$ New speed of train $= s_2 = 60$ kmph