THere are 2 cities 4050 apart. A plane leaves one of them, traveling towards the other at 400 miles per hour. At the same time a plane leaves the other city, traveling at an average speed of 410 miles per hour. How long will it take them to meet?
2007-09-24 06:33:54 · 4 answers · asked by higherlovetx 5 in Science & Mathematics ➔ Mathematics
When the planes meet, the total distance they have both travelled will be 4050 miles. Distance = time * speed, so
400t + 410t = 4050. Now solve for t.
2007-09-24 06:39:02 · answer #1 · answered by Mathsorcerer 7 · 2⤊ 0⤋
Plane A is at city A. Plane B is at city B.
0 + 400(x) = 4500 - 410(x)
810x = 4500
x = 5 5/9 hours
2007-09-24 13:40:45 · answer #2 · answered by PMP 5 · 1⤊ 0⤋
Without going into higher math, the planes are approaching each other and 810 miles/hour and should meet in 5 hours.
2007-09-24 13:37:16 · answer #3 · answered by cattbarf 7 · 1⤊ 0⤋
Write out as two simultaneous equations in distance velocity time;\
x=0+400t
y=4050-410t
_______________
x-y=-4050+810t
When they meet x=y; therefore
t = 5 hours ans.
2007-09-24 13:55:33 · answer #4 · answered by jim m 5 · 0⤊ 0⤋