Two planes leave an airport at the same time, one flying east, the other flying west.The eastbound plane travels 100 mph slower. They are 3040mi apart after 4hrs. find the speed of each plane
let speed of slower plane be x mph
then speed of faster plane is x+100 mph
solve for x
4x + 4(x+100) = 3040
I got x=330 what do i do from there. How do i get the speed of each plane from there?
I defined x as the speed of the slower speed
so the slower plane has speed of 330 mph
I defined x+100 as the speed of the faster plane, so
faster plane goes (330+100) or 430 mph
check:
what is 4(330) + 4(430) ?
3040...okay thanks
Let the speed of the westbound plane be...
x mph, then the speed of the other plane would be (x-100)mph. since the planes are going away in oposite direction so we add the distance they travelled in order to get the distance far apart,
4x +4(x-100)=3040
x is 330mph therefore the speed of the westbound plane is (330-100) which is 230 mph
To find the speed of each plane, we can set up a system of equations based on the given information.
Let's assume the speed of the eastbound plane is x mph. Since the westbound plane is traveling 100 mph slower, its speed would be (x - 100) mph.
We know that the distance covered by both planes after 4 hours is 3040 miles.
For the eastbound plane:
Distance = Speed * Time
Distance = x mph * 4 hrs
Distance = 4x miles
For the westbound plane:
Distance = Speed * Time
Distance = (x - 100) mph * 4 hrs
Distance = 4(x - 100) miles
Since the two planes are flying in opposite directions, the sum of their distances equals the total distance between them:
4x + 4(x - 100) = 3040
Now, let's solve this equation to find the value of x.
4x + 4(x - 100) = 3040
4x + 4x - 400 = 3040
8x - 400 = 3040
8x = 3040 + 400
8x = 3440
x = 3440 / 8
x = 430
Therefore, the speed of the eastbound plane is 430 mph, and the speed of the westbound plane is (430 - 100) = 330 mph.