Plane A leaves an airport in Vancouvar at 10:00 am. At 11:00am plane B leaves the same airport on the same caourse and overtakes the plane A in 2 and a half hours. If the average spped of plane B is 300km/h faster than plane A, find the speed of each?
2006-12-30 13:50:12 · 7 answers · asked by Cutie 4 in Science & Mathematics ➔ Mathematics
Let t be the time run by plane A. t-1 is the time run by plane B.
Va = speed of plane A
Vb = speed of plane B
Vb = 300 + Va
t -1 = 2.5 hr
t = 3.5 hr
Balance by the same distance:
Va t = Vb (t+1)
3.5 Va = 2.5(300 + Va)
Solving for Va gives,
Va = 750 km/h
Vb = Va + 300 = 1050 km/h
2006-12-30 14:09:53 · answer #1 · answered by sahsjing 7 · 0⤊ 0⤋
Hi,
The fastest way to answer this is to say that after the second plane takes off, it takes 2 and a half hours to catch up with the other plane because it is going 300 km/h faster. What that means is that (2.5 hours)(300km/h) = 750km is the distance the second plane had to catch up. That also means that the first plane was flying at 750 km/h because it had a 1 hour head start to go the 750 km that the other plane had to cover to catch up. If the first plane was going 750 km.h and the second plane went 300 km/h faster, then it had to go 1050 km.h.
The more standard rate x time = distance way is to say the first plane traveled a total of 3.5 hours before the other plane caught up at an unknown rate of K km/h. That means 3.5K is the distance traveled by the first plane.The second plane only flies 2.5 hours at a rate of K + 300 km/h since it went 300 km/h faster. That means the second plane traveled a distance of 2.5( K + 300). Since at that point both planes had gone the same distance , you can set those 2 expressions equal to each other.
3.5K = 2.5( K + 300)
Simplifying this, you get
3.5K = 2.5K + 750
K = 750 This is the speed of the first plane.
K + 300 = 750 + 300 = 1050 for the second plane's speed.
I hope that helps.
2006-12-30 14:09:44 · answer #2 · answered by Pi R Squared 7 · 0⤊ 0⤋
Let´s say that the plane A has a rate of a km/h, and the plane B, a rate of b km/h.
Then b - a = 300 km/h
The plane a travels 3 1/2 hours what the plane B travels in 2 1/2 hours. Let's say that the distance that the planes fly is d.
v = d/t
v is the average speed in km/h, and t is the time, in this case, t is in hours
a = d/3.5
b = d/2.5
So, d= 3.5 a = 2.5 b
Then you can solve this equation's system:
- a + b = 300
3.5 a - 2.5 b = 0
Take care
Anabel
2006-12-30 14:03:05 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Speed of A in km/h = v
Speed of B in km/h = v + 300
Distance from Vancouver to point of overtaking = x (same for both planes)
speed = distance / time
Plane A: v = x/3.5
Plane B: v+300 = x/2.5
You now have two unknown and two equations. Go for it. There are many ways to solve this. They should all give the same answer.
2006-12-30 14:03:53 · answer #4 · answered by Raymond 7 · 0⤊ 0⤋
Alrighty then....
Distance = Rate * Time
So....
Let X = Rate of the First Plane
Let 300 + X = Rate of the Second Plane
When the two planes catch up to each other...their distance is equal....so
X * 3.5 (hours) = (X+300) * 2.5 (hours)
3.5X= 2.5X + 750
X = 750
So First plane is 750 km/h
and Second Plane is 1050 km/h
2006-12-30 14:06:48 · answer #5 · answered by Panky1414 2 · 0⤊ 0⤋
Let x be the speed of plane A and y the speed of plane B. In 2.5 hours, plane B has travelled the same distance that plane A has travelled in 3.5 hours, so:
3.5x = 2.5y, or
3.5x - 2.5y = 0
we also know that Y - x = 300, because plane B is 300 km/h faster. We have two equations and two unknowns.
3.5x - 2.5y = 0
-x + y = 300
multiply the second equation by 3.5 and we have:
3.5x - 2.5y = 0
-3.5x + y = 1050
Add the two equations and get:
0x + 1.5 y = 1050
y= 700
x = 700 - 300 = 400
2006-12-30 14:04:37 · answer #6 · answered by David H 4 · 0⤊ 0⤋
Distance equals rate times time.
2006-12-30 13:57:21 · answer #7 · answered by Anonymous · 0⤊ 0⤋