Question

Two planes left the same airport at the same time traveling in opposite directions. The first plane traveled at an average rate of 30 mph slower than the second plane. After 3 hours, the planes were 3150 miles apart.

Enter an equation that can be used to determine when the planes are 3150 miles apart, where x represents the average rate of the second plane.

Enter the average rate of change of the slower plane.

Answer 1

Time for both planes is: t=3
One planes rate (r) is: x+30
The other's rate (r) is: x
The distance of one plane is: rt=tr=3(x+30)
The distance of the other is: rt=tr=3x
Since they're going in opposite directions, if we add their two distances together you get: 3150
The equation to solve is:
3(x+30)+3x=3150
3x+90+3x=3150
6x+90=3150


6x+90-90=3150-90
6x=3060
6x/6=3060/6
x=510

One plane flew at: x+30=510+30=540 mph
The other flew at: x=510 mph

Check if both planes flew at that rate for 3 hours would they be 3150 miles apart?
3(540)+3(510)=3150
1620+1530=3150