Two planes start from the same airport and fly in opposite directions.?

Question

THe first plane is flying 40 mph faster than the second plane. In 5 hours the planes are 2000 miles aprart. Find the rate of the slower plane.

Answer 1

in 5 hours they are 2000 miles apart. so in one hour they have to be 400miles apart.
x>y, y=x-40
x+y=400
2x-40=400
x=220miles per hour
the slower plane goes at 220-40=180mph!

Answer 2

use the rt = d formula to work this problem.

Let r = rate of the slower plane.
r + 40 = rate of the faster plane
5 hours = time the planes fly
2000 = total distance for the two planes.

5r = distance of the slower plane
5(r + 40) = distance the faster plane

5r + 5(r + 40) = 2000
5r + 5r + 200 = 2000
10r + 200 = 2000
10r = 1800
r = 180 mph

Answer 3

If the slower plane is traveling at a speed v, then the faster one is traveling at v + 40 in the opposite direction. Each hour, the distance between them increases by the sum of those speeds, or 2v + 40. So we have

5(2v + 40) = 2000
10v + 200 = 2000
10v = 1800
v = 180, the speed of the slower plane.

Answer 4

Plane 1: speed = (x + 40) mph
Plane 2: speed = x mph
D1 = 5.(x + 40) = 5x + 200
D2 = 5x
5x + 5x + 200 = 2000
10x = 1800
x = 180
Rate of slower plane = 180 mph

Answer 5

the slower plane is 180mph, the faster one is 220mph, their difference is 4omph.

First divide 2000 by 5, so it'll be 400 mph.

You mentioned their difference of 40mph, so let x be the slower plane, and x+40 be the faster plane.

In an hour, you have:
x + (x+40) = 400
2x = 400 - 40
x = 180

So, x (slower) = 180 mph
x + 40 (faster) = 220 mph

Answer 6

let v be velocity of slower plane
w - velocity of faster plane

we're told that

w = v + 40, and
(w+v) * 5 = 2000

substitute w = v+40
(v+40+v)*5=2000
2v+40=400
2v=360
v=180
then, w=180+40=220