Could anyone solve this problem?

A plane flies 450 miles with the wind and 300 miles against the wind in the same length of time. If the speed of the wind is 20mph, what's the speed of the plane in still air?

a) 90mph
b) 105mph
c) 100mph
d) 125mph

Is there a formula to solve this? I am just a student pilot, but it seens to me like I would need the time and heading to solve this!? Am I wrong?
Thank you guys.

My stab at it.

100mph


t(x+20) = 450
t(x-20) = 300

450/(x+20) = 300/(x-20)

(x-20)/(x+20) = 300/450 = 2/3

3x-60 = 2x+40

x = 100

To check ----- 450 miles / 120 mph = 3.75 hours
300 miles / 80 mph = 3.75 hours

pilot27usa wrote:

A plane flies 450 miles with the wind and 300 miles against the wind in
the same length of time. If the speed of the wind is 20mph, what's the
speed of the plane in still air?

a) 90mph
b) 105mph
c) 100mph
d) 125mph

Is there a formula to solve this? I am just a student pilot, but it
seens to me like I would need the time and heading to solve this!? Am I
wrong?
Thank you guys.

Sounds more like a high school algebra problem, but...

distance = speed X time

D = S * t

Let:

Sp = plane speed
Sw = wind speed

therefo

450 = (Sp + Sw) * t
300 = (Sp - Sw) * t

rearranging:

t = 450 / (Sp + Sw)
t = 300 / (Sp - Sw)

eliminate t:

450 / (Sp + Sw) = 300 / (Sp - Sw)

Plug in Sw and solve for Sp.



--
Jim Pennino

Remove .spam.sux to reply.

pilot27usa wrote:
A plane flies 450 miles with the wind and 300 miles against the wind in
the same length of time. If the speed of the wind is 20mph, what's the
speed of the plane in still air?

a) 90mph
b) 105mph
c) 100mph
d) 125mph

Is there a formula to solve this?

Yes. You start by setting up two "distance = speed x time" formulas with
appropriate variables and then combine them to solve for the speed of
aircraft.

I am just a student pilot, but it
seens to me like I would need the time and heading to solve this!? Am I
wrong?

Actually the time factor is the variable common to both distance/time
equations that you use to join the two - in the process the time variable
vanishes.

Not sure why you think you need heading for this problem.

I suppose if your algebra is really rusty you'd have problems composing a
solution.

On May 20, 10:49*am, pilot27usa pilot27usa.
wrote:
A plane flies 450 miles with the wind and 300 miles against the wind in
the same length of time. If the speed of the wind is 20mph, what's the
speed of the plane in still air?

a) 90mph
b) 105mph
c) 100mph
d) 125mph

Is there a formula to solve this? I am just a student pilot, but it
seens to me like I would need the time and heading to solve this!? Am I
wrong?
Thank you guys.

Probably not the best forum, because a pilot's solution (especially
were this to pop up on a written exam) would probably be to run each
of the numbers through an E6B and eliminate each option, which is
quicker and easier to do on a mechanical E6B than it is to figure out
the formula on the fly, especially since it's not particularly
relevant to flight planning. This is a math question. :

In article ,
pilot27usa wrote:

A plane flies 450 miles with the wind and 300 miles against the wind in
the same length of time. If the speed of the wind is 20mph, what's the
speed of the plane in still air?

a) 90mph
b) 105mph
c) 100mph
d) 125mph

Is there a formula to solve this? I am just a student pilot, but it
seens to me like I would need the time and heading to solve this!? Am I
wrong?
Thank you guys.

Why do you think you need the heading? You're told that the plane is
going with/against the wind, i.e. the wind direction and the direction
of travel are the same/opposite. Why do you think you need to know what
that direction actually is?

rg

pilot27usa wrote:
A plane flies 450 miles with the wind and 300 miles against the wind in
the same length of time. If the speed of the wind is 20mph, what's the
speed of the plane in still air?

a) 90mph
b) 105mph
c) 100mph
d) 125mph

Is there a formula to solve this? I am just a student pilot, but it
seens to me like I would need the time and heading to solve this!? Am I
wrong?
Thank you guys.

There is a formula. Here it is; speed times time = distance.
So use x for the unknown still-air speed, t for the time.
Then write out the info:
(x+20)t = 450
(x-20)t = 300

This is soluble using the method called "simultaneous equations".
Or enter it in a smart search engine like alpha, like this:
solve (x+20)t = 450, (x-20)t = 300
Answer: t = 15/4 x = 100
Here's the URL:
http://www98.wolframalpha.com/input/...x-20)t+%3D+300
or in tiny format:
http://tinyurl.com/rb3wno

Brian W