Plane A leaves the airport. One hour later, Plane B leaves the same airport on the same course. It catches up to Plane A in 2.5 hours. The average speed of Plane B is 300km/h faster than Plane A. Find the speed of each plane.

Answer 1

Recall that speed is distance traveled over time:
v = d/t
Represent the speeds with variables:
Let x = speed of Plane A
Since it was said the speed of Plane B is 300 km/h faster than Plane A,
Let x + 300 = speed of Plane B

When they catch up, that means that their distance traveled is the same, so we will equate their distance traveled.
v = d/t
d = v*t

Plane A:
d = x * (2.5 + 1)
Note that the total time traveled by Plane A is 2.5 + 1 because it is one hour ahead of Plane B, and Plane B caught up with it after 2.5 hours.

Plane B:
d = (x + 300) * 2.5

Equating,
x * (2.5 + 1) = (x + 300) * 2.5
3.5x = 2.5x + 750
3.5 x - 2.5x = 750
x = 750 km/h (Plane A speed)
x+300 = 1050 km/h (Plane B speed)

hope this helps~ `u`

Answer 2

Thanks so much! This helped a lot