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 December 14th, 2012, 05:55 PM #1 Member   Joined: Dec 2012 Posts: 30 Thanks: 0 Word Problem An airplane travels 2436 km against the wind in 4 hours and 2996 km with the wind in the same amount of time( 4 hours). What is the rate of the plane in still air and what is the rate of the wind?
 December 14th, 2012, 06:21 PM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Re: Word Problem We have two equations and two unknowns. If we assign w as the wind speed and a as the airplane's speed, then the first part of the problem statement yields the equation: $4(a-w)=2436$ Can you use the second part of the problem statement to come up with the second equation and solve? Please reply if you get stuck.
 December 14th, 2012, 06:33 PM #3 Member   Joined: Dec 2012 Posts: 30 Thanks: 0 Re: Word Problem Do I take both equations and set them equal to each other?
December 14th, 2012, 06:47 PM   #4
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Re: Word Problem

Quote:
 Originally Posted by tallbabe1 What is the rate of the plane in still air and what is the rate of the wind?
Does this mean that the answer would be a fraction because it says what is the rate?

 December 14th, 2012, 06:52 PM #5 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Re: Word Problem In this case I would add the equations together to eliminate w as follows: $4(a-w)=2436 \$ divide by 4: $a-w=609 \$ (1) The other equation is: $4(a+w)=2996 \$ divide by 4: $a+w=749 \$ (2) Adding (1) plus (2) gives: $a-w+a+w=609+749\$ so $2a=1358 \$ and $\ a=679$ Now you can solve either equation for w.
December 14th, 2012, 06:59 PM   #6
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Re: Word Problem

Quote:
Originally Posted by tallbabe1
Quote:
 Originally Posted by tallbabe1 What is the rate of the plane in still air and what is the rate of the wind?
Does this mean that the answer would be a fraction because it says what is the rate?
Unfortunately I am leaving out units, which I should not really do. But I was trying to keep from cluttering up the equations. The units used are km and hr, so the speeds are in $\ \frac{km}{hr} \$ which is consistent with the equations. For example, when we divided by 4 in the first equation we were dividing:

$\frac{2436 \ km}{4 \ hr}= 609 \ \frac{km}{hr}$

December 14th, 2012, 07:00 PM   #7
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Re: Word Problem

Quote:
Originally Posted by tallbabe1
Quote:
 Originally Posted by tallbabe1 What is the rate of the plane in still air and what is the rate of the wind?
Does this mean that the answer would be a fraction because it says what is the rate?
Rate is same as speed...don't try complicating things

 December 14th, 2012, 07:13 PM #8 Member   Joined: Dec 2012 Posts: 30 Thanks: 0 Re: Word Problem So I did the first equation: 4(679-w)=2436 2716-4w=2436 -4w=-280 w=70 wind 70 Speed 679 So simplified is 10/97 Am I doing this right?
 December 14th, 2012, 07:20 PM #9 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Re: Word Problem Well, you worked it correctly except at the end. The airplane, a and wind, w speeds are: $a=679 \ \frac{km}{hr} \$ and $w=70 \ \frac{km}{hr} \$ You should not have taken the ratio. Give me a little time and I will make a follow up post with the full solution including units.
 December 14th, 2012, 07:44 PM #10 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Re: Word Problem Here is the full solution with units: Let a be the airplane's speed relative to the ground in km/hr. Let w be the wind speed relative to the ground in km/hr. So: $4hr \cdot (a-w)=2436km \$ and dividing by 4 hr gives: $a-w=\frac{2436}{4} \cdot \frac{km}{hr} \$ so $a-w=609 \cdot \frac{km}{hr} \$ (1) Notice that equation (1) is dimensionally consistent: a and w are both in km/hr (and so is their difference) on the left side and 609 is in km/hr on the right side. We also have: $4hr \cdot (a+w)=2996km \$ and dividing by 4 hr gives: $a+w=\frac{2996}{4} \cdot \frac{km}{hr} \$ so $a+w=749 \cdot \frac{km}{hr} \$ (2) Equation (2) is also dimensionally consistent: a and w are both in km/hr (and so is their sum) on the left side and 749 is in km/hr on the right side. Adding equation (1) to equation (2) we have: $a-w+a+w=609 \cdot \frac{km}{hr}+749 \cdot \frac{km}{hr}$ $2a=1358 \cdot \frac{km}{hr} \$ so $\fbox{a=679 \cdot \frac{km}{hr}}$ Plugging this value into equation (2) - note that I am using this equation because it is slightly easier to present here: $679 \cdot \frac{km}{hr}+w=749 \cdot \frac{km}{hr} \$ so $w=749 \cdot \frac{km}{hr} -679 \cdot \frac{km}{hr}$ $\fbox{w=70 \cdot \frac{km}{hr}}$

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