Two Planes just took off from Reno, NV. The first plane is traveling 3.5 times as fast as the second plane. After traveling in the same direction for 9 hours, they are 2,070 miles apart. What is the average speed of each plane? (Hint: Since they are traveling in the same direction, the difference between them will be the difference of their distances.)
I came up with plane 1- 141.75 and plane 2- 40.5. I am unsure if this is correct.
Ok, the first step you should take is to define a variable (or variables) to represent unknown quantities in the problem.
You do not know the speed of the slower plane. So,
let x = speed in miles per hour for the slower plane
You don't know the speed of the faster plane either, but you DO know that it is flying 3.5 times as fast as the slower plane. So, if the speed of the slower plane is x mph, then
3.5*x = speed of faster plane in miles per hour.
Each plane flies for 9 hours, based on what it says in the problem.
A VERY IMPORTANT CONCEPT is this one:
distance traveled = rate * time
The time for each plane is 9 hours.
The slower plane traveled 9 hours at x mph. So, distance for the slower plane is 9*x miles.
The faster plane traveled 9 hours at 3.5x mph. So, the distance for the faster plane is 9*(3.5x) miles.
Now...the two planes started at the same time, in
the same place, and traveled in the same direction. We could draw this diagram:
Slow plane: ----------9x miles-------------->
Fast plane: ----------9(3.5x) mi---------------------------->
If you SUBTRACT the distance the slow plane traveled from the distance the fast plane traveled, you'll get how far apart the two planes are at the end of 9 hours
distance between two planes = 9(3.5x) - 9x
AND...we know that they are 2070 miles apart after 9 hours....so,
2070 = 9(3.5x) - 9x
Solve that for x...
Remember that x is the speed of the slow plane. And you know that the speed of the fast plane is 3.5x, so once you know x, you can easily find 3.5x.
