Passenger 18

Passenger and freight trains are operated between stations I and II, where the distance between stations is 16 km. The speed of the passenger trains is 100km/h, and the speed of the freight trains is 60km/h for a single-line railway track. The waiting time for passenger trains is 9 minutes, and 2 minutes for freight trains. It is possible to operate 2 freight trains per day, and the daily operation time is 21 hours.

a) Calculate the number of passenger trains operated daily per direction.
b) If the speed of the passenger trains increases by 30%, find the increase in the number of passenger trains that will be operated daily per direction.

Correct answer:

a =  65
b =  74

Step-by-step explanation:

$$\begin{gathered} s=16\text{km} \\ {v}_1=100\text{km/h} \\ {v}_2=60\text{km/h} \\ {w}_1=9\text{min}\to \text{h}=9:60\text{h}=0.15\text{h} \\ {w}_2=2\text{min}\to \text{h}=2:60\text{h}=0.03333\text{h} \\ {t}_1=s/v_1=16/100=\frac{4}{25}=0.16\text{h} \\ {t}_2=s/v_2=16/60=\frac{4}{15}\doteq 0.2667\text{h} \\ T=21\text{h} \\ {a}_0=\frac{T-2\cdot (t_2+w_2)}{t_1+w_1}=\frac{21-2\cdot (0.2667+0.0333)}{0.16+0.15}=\frac{2040}{31}=65\frac{25}{31}\doteq 65.8065 \\ a=\lfloor a_0\rfloor =\lfloor 65.8065\rfloor =65 \end{gathered}$$

$$\begin{gathered} r=100\%+30\%=1+\frac{30}{100}=1.3 \\ {v}_3=v_1\cdot r=100\cdot 1.3=130\text{km/h} \\ {t}_3=s/v_3=16/130=\frac{8}{65}\doteq 0.1231\text{h} \\ {b}_0=\frac{T-2\cdot (t_2+w_2)}{t_3+w_1}=\frac{21-2\cdot (0.2667+0.0333)}{0.1231+0.15}=\frac{5304}{71}=74\frac{50}{71}\doteq 74.7042 \\ b=\lfloor b_0\rfloor =\lfloor 74.7042\rfloor =74 \end{gathered}$$

Try another example