SOLUTION: Two planes are 1500 mi apart and are traveling toward each other. One plane is traveling 120 mph faster than the other plane. The planes meet in 1.5 h. Find the speed of each plane

Question 422865: Two planes are 1500 mi apart and are traveling toward each other. One plane is traveling 120 mph faster than the other plane. The planes meet in 1.5 h. Find the speed of each plane.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
The distance each plane travels will be given by its speed multiplied by the time until impact.
So we have d1 = v1*t and d2 = v2*t, for planes 1 and 2
We know that the total distance d = d1 + d2, and v2 = v1 + 120
So the system of equations is:
1) d1 = v1*t
2) d2 = (v1 + 120)*t
3) d1 + d2 = 1500
Adding 1) and 2) gives d1 + d2 = v1*t + (v1 + 120)*t
Equating the above equation to the RHS of 3) gives v1*t + (v1 + 120)*t = 1500
Since t = 1.5 h, we have one unknown and we can solve for v1:
1.5*v1 + 1.5*v1 + 180 = 1500 -> v1 = (1500-180)/3 = 440 mph
So the speed of the other plane is v2 = 440 + 120 = 560 mph
Check: 440 mph * 1.5 h + 560 mph * 1.5 h = 660 + 840 = 1500 miles
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