Finding Groundspeed Using Linear Equations

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Homework Help Overview

The problem involves determining the speed of an airplane in still air, given its travel times and distances against and with the wind. The subject area includes linear equations and concepts related to relative motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss setting up equations based on the distance traveled by the airplane against and with the wind. There are attempts to clarify the definitions of variables, particularly distinguishing between the speed of the plane and the speed of the wind. Some participants question the correctness of their calculations and the implications of negative values in their results.

Discussion Status

The discussion is ongoing, with participants exploring various equations and checking their work. Some guidance has been provided regarding the setup of equations, but there is no explicit consensus on the final values or steps needed to arrive at the solution.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for solving the problem. There is also a focus on ensuring clarity in the definitions of variables used in the equations.

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Homework Statement


At maximum speed, an airplane travels 2400 miles against the wind in 6hrs. Flying with the wind, the plane can travel the same distance in 5hrs.
Let x be the maximum speed of the plane and y be the speed of the wind. What is the speed of the plane with no wind.

Homework Equations


well first of all, I need to come up with an equation of inequality or linear for this problem.

The Attempt at a Solution


x= max speed
y=speed of the plane
known:2400m/6hrs against wind
2400m/5hrs with wind

can someone give a hint of an equation to approach this problem?
 
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yuuri14 said:

Homework Statement


At maximum speed, an airplane travels 2400 miles against the wind in 6hrs. Flying with the wind, the plane can travel the same distance in 5hrs.
Let x be the maximum speed of the plane and y be the speed of the wind. What is the speed of the plane with no wind.

Homework Equations


well first of all, I need to come up with an equation of inequality or linear for this problem.

The Attempt at a Solution


x= max speed
y=speed of the plane
known:2400m/6hrs against wind
2400m/5hrs with wind

can someone give a hint of an equation to approach this problem?

y is not the speed of the plane, it is the speed of the wind.

The max speed of the plane is x, which is when the plane's inherent speed (call it v) is added to the speed of the wind [STRIKE](call it w)[/STRIKE] (call it y). Now the one constant thing you have in this problem is the distance that is flown (it's the same both ways, right?) -- call that d.

So write the two equations for how long it takes to travel the distance d, one with the tail wind in the shorter time, and one with the headwind in the longer time.
 
Last edited:
berkeman said:
y is not the speed of the plane, it is the speed of the wind.

The max speed of the plane is x, which is when the plane's inherent speed (call it v) is added to the speed of the wind (call it w). Now the one constant thing you have in this problem is the distance that is flown (it's the same both ways, right?) -- call that d.

So write the two equations for how long it takes to travel the distance d, one with the tail wind in the shorter time, and one with the headwind in the longer time.


So would it rate x time= distance
6(x-y)=2400 against the wind
5(x+y)=2400 with the wind

6x-6y=2400
x-y=400
y=400-x
-------------------------
5x+5y=2400
x+y=480
x-(400-x)=480
2x-400=480
2x=880
x=440
---------------------------
6(x-y)=2400
6(480-y)=2400
2880+2y=2400
2y=-480
y=-240
is this wrong I ended up w/ a (-) number
 
6(480-y)=2400
2880+2y=2400

Re-check this step. You were correct up to that point. Good job!
 
berkeman said:
Re-check this step. You were correct up to that point. Good job!

opps my mistake.
6(x-y)=2400
6(480-y)=2400
2880-6y=2400
-6y=-480
y=80
is this right. So is this the answer if their no wind for the plane or is there another step to it?
 
yuuri14 said:
opps my mistake.
6(x-y)=2400
6(480-y)=2400
2880-6y=2400
-6y=-480
y=80
is this right. So is this the answer if their no wind for the plane or is there another step to it?

x is not 480. Double check what you got above when you solved for x.

And x is the max speed of the plane, right? And y is the speed of the wind. So what is the speed of the plane with no wind?
 
berkeman said:
x is not 480. Double check what you got above when you solved for x.

And x is the max speed of the plane, right? And y is the speed of the wind. So what is the speed of the plane with no wind?

6(x-y)=2400
6(440-y)=2400
2640-6y=2400
-6y=-240
y=40

well i know that:
x is the max speed of the plane :440
y is the speed of the wind:40
if i divide it its: 11...is that the answer?
 
yuuri14 said:
6(x-y)=2400
6(440-y)=2400
2640-6y=2400
-6y=-240
y=40

well i know that:
x is the max speed of the plane :440
y is the speed of the wind:40
if i divide it its: 11...is that the answer?

Don't divide. Don't guess.

If you have an airspeed of 100mph, and a 10mph tailwind, what is your groundspeed? Think about it...
 
berkeman said:
Don't divide. Don't guess.

If you have an airspeed of 100mph, and a 10mph tailwind, what is your groundspeed? Think about it...

why do i have to use 100mph, and a 10mph tailwind? It wasn't given to the original problem.
 

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