Finding Groundspeed Using Linear Equations

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In summary: Why?Why do you have to use 100mph, and a 10mph tailwind? It wasn't given to the original problem. Why?
  • #1
yuuri14
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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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  • #2
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:
  • #3
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
 
  • #4
6(480-y)=2400
2880+2y=2400

Re-check this step. You were correct up to that point. Good job!
 
  • #5
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?
 
  • #6
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?
 
  • #7
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?
 
  • #8
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...
 
  • #9
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.
 

1. What is linear problem solving?

Linear problem solving is a method used in mathematics and science to find the solution to a problem by representing it in a linear equation or system of equations. It involves using variables and constants to represent the unknowns and knowns in the problem, and then using algebraic techniques to solve for the variables.

2. When should linear problem solving be used?

Linear problem solving is most useful when the relationship between the variables in the problem is linear, meaning that they can be represented by a straight line on a graph. It is also helpful when the problem involves multiple variables and their interactions, as it allows for a systematic approach to finding a solution.

3. What are the steps to solving a linear problem?

The first step is to identify the variables and constants in the problem, and assign them symbols. Then, write out the equations that represent the problem using these symbols. Next, use algebraic techniques such as substitution, elimination, or graphing to solve for the variables. Finally, check your solution to make sure it satisfies all of the equations in the problem.

4. Can linear problem solving be applied to real-world situations?

Yes, linear problem solving is commonly used in various fields such as engineering, economics, and physics to model and solve real-world problems. It can be used to analyze and optimize systems, predict outcomes, and make decisions based on data.

5. Are there any limitations to using linear problem solving?

Linear problem solving is limited to problems with linear relationships between variables. If the problem involves non-linear relationships, other methods such as quadratic or exponential equations may need to be used. Additionally, it assumes that all variables are continuous and can take on any value, which may not always be the case in real-world situations.

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