Weird trig problem?

Question

I haven't the foggiest idea how to even START this problem. You don't necessarily have to give me the answer, but could you maybe walk me through how I would go about solving it?

"A plane flies 600 miles with a tail wind in 2 hours. It takes the same plane 4 hours to fly the 600 miles when flying against the wind. What is the plane’s speed in still air?"

Answer 1

Don't see why trig is involved, frankly.

w = wind speed (mph)
p = plane speed (mph).
t = time (h)
d = distance (miles)

t * (p+w) = d

2 * (p+w)=600; (p+w)=300; p+w-300=0
4 * (p-w) = 600; (p-w) = 150; p-w-150=0

p+w-300=p-w-150
2w=150
w = 75
p = 225

So the plane flies 225 in still air. Going with a tail wind, he travels 300 mph, or 600 miles in 2 hours. Going into a head wind, he travels 150 mph, or 600 miles in 4 hours.

This is straightforward algebra. Perhaps the problems author is trying a change-up on you to see if you're awake?

Answer 2

Assuming the wind speed is the same in both cases, and that the wind velocity (a vector that has speed and direction) is parallel to the velocity vector of the plane (i.e. in the same direction of directly opposite).

So if the trip is from A to B, the distance being 600 miles, and the wind speed = Vw, and plane speed in still air Vp

In the first trip the wind adds to the planes velocity so the velocity Vab = + Vw

So the trip time from A to B, = (Vp+Vw)/600 = 2

So Vp + Vw = 1200

In the return journey the wind acts against the plane

And the trip from B to A, = (Vp-Vw)/600 = 4

So Vp - Vw = 2400

Adding both these equations together, the wind speed cancels out

So 2Vp = 3600

And

Vp = 1800 miles/hr

Answer 3

I believe that the problem assumes that the airplane travels at a particular speed relative to the wind. I believe that the problem also assumes that the wind was constant throughout both trips.

The speed of the airplane relative to the ground in the first trip was 300 miles per hour

The speed of the airplane relative to the ground in the second trip was 150 miles per hour.

If the wind speed was indeed constant then the speed of the wind was half the difference between these two numbers. Subtracting the wind speed from the larger speed should be the speed relative to the ground of the airplane in still air.


A second way to get the same answer.
The speed of the airplane in still air would be the average of the two ground speeds.

Answer 4

Air speed with a tail wind is 300 mph
Air speed against the wind is 150 mph

total speed 450 mph

2 trips then divide 450 by 2 = 225 mph

Answer 5

This is just an algebra problem. No trig involved.

Let

p = speed of plane in still air
w = speed of wind

p + w = speed of plane with tail wind
p - w = speed of plane with head wind

Remember the formula

distance = rate x time

2(p + w) = 600
4(p - w) = 600

p + w = 300
p - w = 150
2p = 450
p = 225 miles per hour

Answer 6

well, distance is rate * time
and the velocity is either stillair + the windspeed, or stillair - windspeed
i'll call still air speed p, and windspeed w
so 600 = 2*(p+w)
and 600 = 4*(p-w)
solve the first equation for w:
300 - p = w
substitute into the second equation and solve for p
600 = 4*(p - (300 - p))
600 = 4*(2p - 300)
600 = 8p - 1200
1800 = 8p
p = 225 mph
and that's your answer

Answer 7

Let us assume the plane speed is x mile and wind speed is y

so use distance=inital speed * time

2*(x+y)=600

4*(x-y)=600

solve for x

the answer is x=225 miles per hour is the plane speed in still air

Answer 8

with a tail wind it goes 300 mph and with the headwind it goes 150 mph. that means the wind is blowing at 75 mph and the speed in no wind would be 225 mph. i dont think theres any trig there at all.

Answer 9

let x=airspeed & y= wind speed
600/(x+y)=2
300=x+y

600/(x-y)=4
150=x-y
300=x+y add
450=2x
x=225
airplane speed in still air is 225 mph
x+y=300
y=300-x=300-225
y=75
wind speed=75 mph

Answer 10

The answer is 237 miles/hr. I tried pasting the solution but am unable to do that. Is there anyway I can email the solution to you?

Nelly