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Mathematics 13 Online
OpenStudy (anonymous):

An airplane flying with a tail wind can complete a journey of 3500 kilometers in5 hours. flying the reserve direction, the plane completes the same trip in 7 hours. what is the speed of the plane in still air??

OpenStudy (anonymous):

Reverse direction*

OpenStudy (anonymous):

Think about it this way: There are basically two variables. We don't know the speed of the plane without wind, and we don't know the speed of the wind. Let's call the plane's normal speed S Let's call the speed of the wind W When the wind is moving with the plane, it makes the plane go faster, so the speed will be S+W When the wind is moving against the plane, it slows it down, and the speed will be S-W

OpenStudy (anonymous):

Now, you should know that \[\frac{\text{distance}}{\text{speed}} = \text{time}\] So use that two write two equations representing the two trips.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Can I see your equations?

OpenStudy (anonymous):

well im stuck that is why i wanted to see if someone can show me step by step, I know that the answer is 600 kilometers per hour, but idk how to get there

OpenStudy (anonymous):

Writing the two equations is step 1 =)

OpenStudy (anonymous):

ok can you hold on for a min

OpenStudy (anonymous):

Walk through this with me. You can write these two equations. \[\frac{\text{distance}}{\text{speed}} = \text{time}\]

OpenStudy (anonymous):

Gladly =)

OpenStudy (anonymous):

ok so the first step is find distance/speed= time?

OpenStudy (anonymous):

?

OpenStudy (anonymous):

are u there?

OpenStudy (anonymous):

To write an EQUATION by using that fact. We have two situations the trip there and the trip back. Start with the trip there. Figure out what to put in as the distance, the speed, and the time.

OpenStudy (anonymous):

time= 5

OpenStudy (anonymous):

distance and speed idk

OpenStudy (anonymous):

???

OpenStudy (anonymous):

lol maybe..goshh i really need help :(

OpenStudy (shane_b):

You need to come up with 2 equations. You're given enough information to complete the first equation which is for the first trip:\[Speed=\frac{distance}{time}=\frac{?}{?}\]

OpenStudy (anonymous):

yeah, is time 5 hours?

OpenStudy (shane_b):

Yes. @colombianita19 : sorry for butting in...thought you left. Please continue :)

OpenStudy (shane_b):

err I meant @SmoothMath

OpenStudy (anonymous):

did he leave?

OpenStudy (shane_b):

It looks like he's in the room but maybe afk.

OpenStudy (shane_b):

So what does your first equation look like?

OpenStudy (anonymous):

is it speed= distance/ time

OpenStudy (shane_b):

Yes...plug in the values and solve for the speed. That's the first part.

OpenStudy (anonymous):

Distance is also given... if you read the problem.

OpenStudy (anonymous):

is the value for time= 5hours?

OpenStudy (anonymous):

And for speed, go back and read my first post.

OpenStudy (shane_b):

yes

OpenStudy (shane_b):

<backing out since Smooth is back>

OpenStudy (anonymous):

distance is 3500 kilometers?

OpenStudy (anonymous):

for speed is just s?

OpenStudy (anonymous):

Yes, 3500km, and that's the same for both situations, right?

OpenStudy (anonymous):

Shane, go ahead and take over. I'm distracted.

OpenStudy (anonymous):

there and back?

OpenStudy (shane_b):

Ok...here's your first equation:\[Speed=\frac{3500km}{5hr}\]So what's the speed?

OpenStudy (anonymous):

is it just s? doesn't say

OpenStudy (shane_b):

Look at what I just wrote and solve for the speed :)

OpenStudy (anonymous):

700?

OpenStudy (shane_b):

Yes... So during the first trip the plane with the tailwind traveled at 700km/hr.

OpenStudy (shane_b):

Now for the return trip. What do you think the formula will look like for that trip?

OpenStudy (anonymous):

time/ distance?

OpenStudy (shane_b):

No...it's the same formula: \[Speed = \frac{distance}{time}\]Now plug in what you know for the return trip: It has the same distance which is 3500 km...and it took them 7 hours to return. So what's the speed on that leg of the trip?

OpenStudy (shane_b):

<The goal here is to find the difference in speed between both legs of the trip...so you find the speed on each leg and then calculate the difference>

OpenStudy (anonymous):

is speed 500?

OpenStudy (shane_b):

Yes. So they went 700 km/hr to get there...WITH a tailwind...and they went 500 km/hr to return WITHOUT a tail wind. So what was the speed of the tailwind?

OpenStudy (anonymous):

600?

OpenStudy (anonymous):

200

OpenStudy (shane_b):

Oops...just noticed I made a silly mistake. You were sort of right. The speed of the plane in still air is 600 km/hr. When it has a 100 km/hr tailwind it goes 700 km/hr...and when it goes back the other way it has a 100 km/hr headwind...which slows it down to 500 km/hr.

OpenStudy (anonymous):

on the back of the book is said that the answer is 600 kilometers per hour

OpenStudy (shane_b):

600 is right...see what I just wrote.

OpenStudy (anonymous):

so what part of the equation did we make the mistake

OpenStudy (anonymous):

trip there i have 350/ 5 hrs speed = 700....and trip back i have 3500/7hrs= 500

OpenStudy (anonymous):

opsss i mean 3500 on both equations

OpenStudy (shane_b):

MY mistake was forgetting about the headwind on the way back. What this problem boils down to is calculating the difference in speed which I hope you know how to do now. There was a 200 km/hr difference in speed between flying with a tailwind and flying against a headwind. The speed of the plane will actually be in the middle....600 km/hr.

OpenStudy (anonymous):

ok thanks! :)

OpenStudy (shane_b):

I hope I didn't totally confuse you :) your welcome

OpenStudy (shane_b):

*you're

OpenStudy (anonymous):

no you are fine! :)

OpenStudy (anonymous):

Thanks for askin' before takin' over, Shane. Totally considerate of you. You're a champ.

OpenStudy (anonymous):

thanks for trying to help smoothmath:)

OpenStudy (anonymous):

Aw, you're sweet. My pleasure =)

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