Two airplanes leave Dallas, Texas traveling in opposite directions. One is traveling 28 mph faster than the other airplane. How fast is the faster airplane traveling if the two airplanes are 2,320 miles apart after 2 hours and 30 minutes?
f - s = 28
2.5 (f + s) = 2320
f + s = 928
adding equations to eliminate s
... 2 f = 956
To find the speed of the faster airplane, we need to first find the combined speed of both airplanes.
Let's assume the speed of the slower airplane is "x" mph.
Since the faster airplane is traveling 28 mph faster than the slower airplane, its speed is "x + 28" mph.
Now, we know that distance = speed × time.
The slower airplane has been flying for 2 hours and 30 minutes, which is equivalent to 2.5 hours. So, the distance traveled by the slower airplane is 2.5x miles.
The faster airplane has also been flying for 2 hours and 30 minutes, so its distance traveled is 2.5(x + 28) miles.
Since the two airplanes are traveling in opposite directions, their distances traveled will add up to the total distance between them. In this case, the total distance is 2,320 miles.
So, we can write the equation: 2.5x + 2.5(x + 28) = 2,320.
Now, we can solve this equation to find the value of "x" and then calculate the speed of the faster airplane.
2.5x + 2.5x + 70 = 2,320.
5x + 70 = 2,320.
5x = 2,320 - 70.
5x = 2,250.
x = 2,250 / 5.
x = 450.
Now, we know that the speed of the slower airplane is 450 mph.
To find the speed of the faster airplane, we would substitute this value back into the equation: x + 28.
So, the speed of the faster airplane is 450 + 28 = 478 mph.