You are flying your plane. Just as you fly over a building you pass directly underneath the Re/Max hot-air ballon, drifting in the steady wind. 15 minutes later you turn directly around (i.e.180 deg) and fly back. You pass directly underneath the ballon 5km west of the building. What was the speed and direction of the wind?

PS

Any help would be greatly appreciated. Thanx

This isn’t physics…it’s math. Worse…it’s a dreaded WORD problem!

Even though I probably shouldn’t help with your homework I’ll have a stab at it since I’m probably wrong anyway and thus your attempt at cheating will backfire.

The wind direction is due west since the ballon is 5 km due west of the building. You must have been travelling east on the first pass if you expect to turn 180 degrees and pass directly under the balloon again (I’ll ignore the fact that a plane turns in an arc and not in place for now).

The wind’s speed is easy unless you get specific. I doubt a balloon actually travels at the speed of the wind but probably a little slower IRL. Taking an ideal assumption (that the balloon moves at the same speed as the wind) the the wind speed is 20 km/hour. 5 km in 15 minutes --> 15*4=60 minutes (1 hour). 5*4=20km Hence 20 km/hr.

20 km/hr due west.

You know that distance = rate * time, I trust? Or, in this case, rate = distance/time? Then perhaps, given that the problem tells us the distance it travelled, and the time it took to travel, and it did not suggest there was any other propulsion acting on the balloon other than the wind, we can then determine the wind’s speed.

What arlover said – simplify the problem.

The stuff about the airplane is just extraneous information. The important thing to figure out when doing these problems is, what would I have to know to figure out the answer? Keep it simple as possible and work backwards until you have something you know, then you can ignore the rest.

In this case for example :

You need to know the velocity of the wind.

Simplest would be if the problem said, “the wind speed is 20 km/hr, blowing due west”.

Don’t have that, but we know speed is km/hr (distance/time), so if we can get those and the direction, we have the answer.

Is there something that tracks the wind speed? Yes, the balloon. So follow the balloon only; it moved from directly over the building to 5 km west of the building.

How long did it take? 15 minutes. Then using just that, you can get the answer.

::Former High School Physics teacher enters the room::

However, if you pass the balloon, fly for 15 minutes before turning around and then turn around, one should assume that it takes approximately 15 minutes for the return flight. this makes the round trip 30 minutes and not 15.

(To be truly precise, we should know the air speed of the plane because it takes some finite amount of time to fly the 5 km from the first passing of the balloon to the second. Also, we must consider that the airspeed of the plane is not constant as it flies into the wind in one direction and against the wind in the other direction. Further, we don’t know the direction of travel for the plane (E -> W or W -> E), just that it flies along that line.)

I’ll try to code this out.

O = balloon

X = Plane

start

O-------------------------------------------------->X

return

___________OX<---------------------------------------

or

start

___________O--------------------------------------->X

return

OX<--------------------------------------------------

In the first example, we assume that the balloon is travelling west and the plane starts out travelling west. In the second example, we assume that the balloon is travelling west and the plane starts out travelling east.

All this is probably beyond the scope of the question as the change in position of the balloon over time is much less than the change in position of the plane over the same time interval. Further, I doubt any High School introductory physics class would expect this.

The problem is worded poorly. You don’t have enough information to solve it *as written*. You do have enough information to solve the problem the others answered.

See, the plane itself will be subject to the wind, so you would have to know the plane’s airspeed.

Another problem with the way it is written - it says you *turn around* 15 minutes later. According to that, you must travel 15 minutes (or thereabouts) back to the balloon. But it won’t be exactly 15 minutes, only 14 and a half. Without knowing the difference in travel time, your travel time for the balloon is off, so the airspeed of the balloon can’t be calculated correctly.

As for the plane turning around in an arc, true, but we could also assume it corrected for that turn to return to original flight line, as that wouldn’t affect the distance or time significantly.

This problem statement would be better:

“You are flying your plane. Just as you fly over a building you pass directly underneath the Re/Max hot-air ballon, drifting in the steady wind. **A few** minutes later you turn directly around (i.e.180 deg) and fly back. You pass directly underneath the ballon 5km west of the building, **15 minutes after the first pass**. What was the speed and direction of the wind?”

[list=1]

[li]Since the balloon was 5 km west of the building, the wind was blowing from the east to the west.[/li][li]You encounter the balloon after turning around *before* you make it back to the building. Therefore, you were flying west initially, then turned back due east.[/li][li]The speed of your plane is faster going west, since it has a tail wind, and slower going east, since it has a headwind.[/li][/list=1]

Let:

V[sub]p[/sub] = velocity of plane

V[sub]w[/sub] = velocity of wind

t[sub]o[/sub] = time going out from building = 15 min = 1/4 hr

t[sub]i[/sub] = time coming in to building before re-meeting balloon

d = distance plane traveled before turning around

The following equations can be set up:

[list=1]

[li]V[sub]w[/sub](t[sub]o[/sub] + t[sub]i[/sub]) = 5 {velocity of wind times the plane’s total flying time is distance balloon drifted, 5 km}[/li][li]t[sub]i[/sub](V[sub]p[/sub] - V[sub]w[/sub]) = d - 5 {time of plane flying in times its velocity minus headwind is distance d minus the balloons drift distance}[/li][li]t[sub]o[/sub](V[sub]p[/sub] + V[sub]w[/sub]) = d {time of plane flying out times its velocity plus tailwind is distance d}[/li][/list=1]

From {1} equation:

V[sub]w[/sub](1/4 + t[sub]i[/sub]) = 5

V[sub]w[/sub]/4 + V[sub]w[/sub]t[sub]i[/sub] = 5 {4}

Expanding {2}:

V[sub]p[/sub]t[sub]i[/sub] - V[sub]w[/sub]t[sub]i[/sub] = d - 5 {5}

Adding: {4} and {5}

V[sub]p[/sub]t[sub]i[/sub] - V[sub]w[/sub]t[sub]i[/sub] + V[sub]w[/sub]t[sub]i[/sub] + V[sub]w[/sub]/4 = d - 5 + 5

V[sub]p[/sub]t[sub]i[/sub] + V[sub]w[/sub]/4 = d {6}

Expanding {3}:

V[sub]p[/sub]/4 + V[sub]w[/sub]/4 = d {7}

Since both {6} and {7} equal d:

V[sub]p[/sub]t[sub]i[/sub] + V[sub]w[/sub]/4 = V[sub]p[/sub]/4 + V[sub]w[/sub]/4

Subtract V[sub]w[/sub]/4:

V[sub]p[/sub]t[sub]i[/sub] = V[sub]p[/sub]/4

Dividing by V[sub]p[/sub]:

t[sub]i[/sub] = 1/4 hr

Substituting into {1}:

V[sub]w[/sub](1/4 + 1/4) = 5

V[sub]w/sub = 5

V[sub]w[/sub] = 10 km/h, to the west

It matters not what the plane’s speed is. If it were going 90, its net speed out would be 100 km/h and its speed in would be 80. In the 15 minutes out, it’d go 25 km; coming in it’d go 20 km. Difference 5 km.

Or if it were going 250 km/h:

Out: 260 km/h * 1/4 h = 65 km

In: 240 km/h * 1/4 h = 60 km

Net difference: 5 km.

With all due respect to AWB, he has somewhat overcomplicated the solution. I’m guessing that the problem was assigned as part of a unit on relativity (Galilean, not Special).

It is natural for us land-based creatures to focus on the reference frame in which the building is fixed, but that isn’t the reference frame in which the solution is simplest. Note that the balloon is stationary in the reference frame in which the *air* is stationary. Also note that the plane’s motion is also most easily expressed relative to the air. Then, if we assume the plane’s air speed (rather than ground speed is constant), we know immediately that, between the two events described, 30 minutes have passed.

*Now* we go back to the ground frame of reference, where we know the balloon has traveled 5km. Since that took 30 minutes, the wind speed is 10 km/hr.

Thus illustrating the principle that the laws of physics are independent of reference frame, but that the equations are usually simpler in a *particular* frame.

Rick

How do we know it was *before*? Did I miss something in the question?

Now, I’m imagining that it doesn’t matter if the plane is flying parallel to the direction of balloon travel (E-W) or not, since the wind moving the balloon in a westerly direction is also moving the plane **in the same direction**. (map it out if the baloon is moving due west and the plane is travelling due north. However, the round trip time is still 30 minutes!!

Hey, **elontennis**, if you had known that you would have gotten such an indepth response, would you have posted to begin with? I mean you’re doing simple algebra stuff and we’re getting ready to take some major derivatives!!

Misplaced the d*mn parenthesis! Should be,

…Then, if we assume the plane’s air speed (rather than ground speed) is constant,…

Rick (who used preview, but didn’t do it carefully)

Actually, it was remiss of me to assume that the plane encountered the balloon before passing the building again.

However, I tried it the other way (with the plane going east then turning around and catching up with the balloon). Same answer.

Then, thinking intuitively, it doesn’t matter which way the plane is going! If it went north, it’d go V[sub]p[/sub]t[sub]o[/sub] km north and be pushed 2.5 km to the west by the same wind that’s pushing the balloon. Turning around and going south, it’d be pushed another 2.5 km west by the same wind. t[sub]o[/sub] = t[sub]i[/sub], and the plane would be at the same latitude of the building and balloon, and the same longitude as the balloon because of the wind. 5 km in 1/2 hr is 10 km/hr.

I’m in agreement with RickG and AWG’s last response. The answer is 10 km/hr. It’s as simple as recognizing 5 km in half an hour.

It matters not what the direction or speed of the airplane is (as long as it’s greater than zero). The airplane drifts in the same mass of air as the balloon, and its round trip takes half an hour.

Ground-based observers would see that the track of the airplane over the gound would not necessarily be the same as its heading (due to crosswind), or that its groundspeed would generally be different for the out- and inbound legs, but that’s not relevant to this question.

Quoth **RickG**:

…Then, if we assume the plane’s air speed (rather than ground speed) is constant,…

I feel obliged to point out that the problem, as stated, never says that the plane’s speed is constant in *any* particular reference frame. Of course, it’s a reasonable assumption, and the problem’s unsolvable without knowing that (or some other information about the plane’s speed), but it’s still a point.

*Originally posted by Chronos *

**QuothRickG:…Then, if we assume the plane’s air speed (rather than ground speed) is constant,…

I feel obliged to point out that the problem, as stated, never says that the plane’s speed is constant in

anyparticular reference frame. Of course, it’s a reasonable assumption, and the problem’s unsolvable without knowing that (or some other information about the plane’s speed), but it’s still a point. **

Which is pretty much why I made the assumption. I got so used to solving problems like this when I was a Freshman Physics TA that I tend to read the necessary assumptions into the problem without really thinking about it. You are correct, though, that the problem should have stated it explicitly.

Cheers,

Rick