You have the following information:
- Average speed of cattle train to New York: 41.4 mph
- Average speed of passenger train: 22.5 mph
- The passenger train left in the opposite direction, 5.6 hour after cattle train started its travel.
In order to determine how long does the passenger need to travel before the trains are 513 mi apart, you take into account that you can express the previous situation in an algebraic way. If you consider x as the distance traveled by cattle train in a time t, the you have:
x = vt = (41.4)t = 41.4 t
Now, if you consider x' as the distance traveled by the passenger train in the opposite direction in a 5.3h after the left of cattle train, you have:
x' = v't = (22.5)(t + 5.3) = 22.5 t + 119.25
Next, if you are interested in the time on which passengers and cattle train will be separated by 513 mi, then you can write:
x - (-x') = 513 Here, you specify the distance between both trains are 513
x + x' = 513
The minussign of -x' is due to the fact the passengers trains goes in the opposite direction.
Then, by replaacing the expressions for x and x' you obtain:
(41.4t) + (22.5t + 119.25) = 513
Now, you can simplify the previous expression, and solve it for t:
41.4t + 22.5t + 119.25 = 513
63.9t = 513 - 119.25
63.9t = 393.75
t = 393.75/63.9
t = 6.16
Hence, both trains will be at a distance of 513 mi apart between them, after 6.16 hours