At noon a private plane left austin for los angeles 2100km a

At noon a private plane left austin for los angeles, 2100km away, flying at 500km/h. one hour later a jet left los angeles for austin at 700k/m. At what time did they pass each other?

Solution

You will need to apply the distance formula:
d = rt
where
d is distance
r is rate or speed
t is time
.
They will pass each other when the sum of the distance traveled by each plane equals 2100km:
Let t = time they will pass each other
then
500t + 700(t-1) = 2100
500t + 700t - 700 = 2100
500t + 700t = 2800
1200t = 2800
t = 2800/1200
t = 7/3 hour
or
2 hours and 20 minutes
.
Since the time the first plane started was 12 noon,
They will pass each other at 2:20 PM