At noon a private plane left Austin for Los Angeles, 2100 km away, flying at 500km/h. One hour later a jet left Los Angeles for Austin at 700 km/h. At what time did they pass each other?

Answer 1

Let t be the time after the first jet leaves for LA, and x be the location of a plane, measured from LA.

2100 - x = 500 t
x = 700 (t -1) = 700 t - 700
Substitute 2100 - 500 t for x in the second equation and solve for t.

2100 - 500 t = 700 t - 700
2800 = 1200 t
t = 7/3 hours = 2 hours & 20 minutes