Here's a math word problem for Algebra?

Question

An airplane flying into a headwind travels the 1800 mile flying distance between two cities in 3 hours and 36 minutes. On the return flight, the distance traveled in 3 hours. Find the air speed of the plane and the speed of the wind, assuming that both remain constant.

What formula do I use? how to solve? i've got liek.. five other problems like this. Can anyone help me?

Answer 1

Formula to be used
speed = distance/time
Let the airplane speed be x mph
Let wind speed be y mph
When flying into headwind the effective speed =x-y mph
During return flight effective speed = x+y mph
We hence get
x-y=1800/(3+36/60)=1800/3.6=500
x+y=1800/3=600
Adding the two equations
2x=1100
x=550 mph
y=600-550=50 mph
Airplane speed = 550 mph
Wind speed = 50mph

Answer 2

Lets see...

ok... If my math is right:

1800 miles = d
3h36m = 216 min = t
X = Plane speed
Y = Wind speed

First of all, it flew 1800 miles in 216 minutes which is about

1800/216= 8.33333 Miles per minute * 60 mins in an hour = 500MPH

Then you have the return flight, if it made it in 3 hours, and these are the same 1800 miles, the overall speed is 1800/3= 600 MPH

My guess is that if the plane had headwind going, and tailwind returning, you should calculate the difference.

X = 600MPH - Y
X = 500MPH + Y

600 - Y = 500 + Y
600 = 500 + Y + Y
600 - 500 = 2Y
100 = 2Y
Y = 50 this is your wind speed.

X = 600 - 50 = 550
X= 500 + 50 = 550

X = 550 this is your plane speed

Answer 3

1800 miles / 3.6 hours = 500 mph
1800 miles / 3 hours = 600 mph

Speed of the plane = (500 + 600) / 2 =550 mph
Speed of the wind = 50 mph

The wind is subtracted from the air speed (going) and is added to the air speed (return) to give you ground speed each way.

Answer 4

Geez...you seriously have trouble over this?

total distance = 1800miles=d
time=3hr36min=216min=t
So speed whilst heading towards the wind=d/t(60)=500mph
Speed while travelling back=d/3=600mph
So wind speed=600-500=100mph

I hope that helps.

Answer 5

you decide directly to be attentive to t whilst -16t²+256 = 0 As stated, to unravel this equation element thoroughly: -sixteen(t² ? sixteen) = 0 Which will become -sixteen(t + 4)(t – 4) = 0 To be equivalent the two t + 4 = 0 or t – 4 = 0 The values of t that resolve the equation are -4 or 4. It can't be detrimental, so it takes 4 seconds for the rock to return and forth 256 feet because it splashes down. ProfRay

Answer 6

speed of wind = w
speed of plane = p
time required to travel first flight = (1800)/(p-w)
3 hrs = 1800/(p-w)
for the second flight
3hrs 36mins = 1800/(p+w)
two equations two unknowns. bread and butter. its all yours