Question

A "Local" train leaves a station and runs at an average rate of 35 mph. An hour and a half later an "Express" train leaves the station and travels at an average rate of 56 mph on a parallel track. How many hours after the Express train starts will the it overtake the Local?

Answer 1

2.5 hrs.

Hope this helps <3

Answer 2

recall your d = rt, distance = rate * time.

so...Local say "L" is going at a speed of 35mph...ok... and Express or "X" is going at 56mph.

by the time the two trains meet, and X is ready to overtake L, the distance that both have travelled, since is a parallel road, is the same, say "d". So if L has travelled "d" miles, then X had travelled "d" miles too, over the same road, maybe different lane.

now, because X left 1 1/2 hour later, by the time they meet, say X has been running for "t" hours, but because it left 1 1/2 hour later, L has been running for " t + 1 1/2 " hours, or " t + 3/2 " hours.


`\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ Local&d&35&t+(3)/(2)\\ Express&d&56&t \end{array} \\\\\\ \begin{cases} \boxed{d}=35\left( t+(3)/(2) \right)\\ d=56t\\ ----------\\ \boxed{35\left( t+(3)/(2) \right)}=56t \end{cases} \\\\\\ 35t+\cfrac{105}{2}=56t\implies \cfrac{105}{2}=21t\implies \cfrac{105}{42}=t \\\\\\ \cfrac{5}{2}=t\implies 2(1)/(2)=t`

so, they met 2 and a half hours later after X left, and a milllisecond later X overtook L.