I can not figure out how to solve this problem:
"Two planes left the Chicago airport at the same time four hours ago. One is 40 miles per hour faster than the other one, and they are 6600 miles apart. What is the rate of each plane?"
I am not the best at distance problems like this. And it's an extra credit problem, which I really need. So can someone please help me?
I know it would be 6600=4r+40 or something...
speed 1 = r
speed 2 = (r+40)
assume opposite directions
6600 = 4 (r + 40) + 4 r
6600 = 4 r + 160 + 4r
8 r = 6440
r = 805
r+40 = 845
To solve this problem, you can use the concept of relative speed. Let's break it down step by step:
1. Let's assume the slower plane is traveling at a certain rate, say 'r' miles per hour. Since the faster plane is 40 miles per hour faster, its rate can be expressed as 'r + 40' miles per hour.
2. Both planes have been flying for four hours, so the distance covered by each plane can be calculated by multiplying the rate by the time. Thus, the slower plane has covered a distance of 4r miles, and the faster plane has covered a distance of 4(r + 40) miles.
3. Since the total distance covered by both planes is 6600 miles, we can write an equation: 4r + 4(r + 40) = 6600.
Simplifying the equation:
4r + 4r + 160 = 6600
8r + 160 = 6600
4. Now, let's isolate 'r' by subtracting 160 from both sides of the equation:
8r + 160 - 160 = 6600 - 160
8r = 6440
5. Divide both sides of the equation by 8 to solve for 'r':
8r/8 = 6440/8
r = 805
So, the slower plane is traveling at 805 miles per hour, and the faster plane is traveling 805 + 40 = 845 miles per hour.
Therefore, the rate of each plane is 805 mph and 845 mph, respectively.