Answer:
Approximately [tex]116\; \text{miles}[/tex] for the train from Boston to NYC Penn Station.
Approximately [tex]105\; \text{miles}[/tex] for the train from NYC Penn Station to Boston.
Explanation:
Convert minutes to hours:
[tex]\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}[/tex].
[tex]\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}[/tex].
Calculate average speed of each train:
[tex]\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}[/tex].
[tex]\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}[/tex]
Assume that it takes a time period of [tex]t[/tex] for the trains to pass by each other after departure. Distance each train travelled would be:
[tex]s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t[/tex].
[tex]s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t[/tex].
Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:
[tex]v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}[/tex].
Rearrange and solve for [tex]t[/tex]:
[tex](v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}[/tex].
[tex]\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}[/tex].
Distance each train travelled in [tex]t = (539 / 279)\; \text{hour}[/tex]:
[tex]\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}[/tex].
[tex]\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}[/tex].
