Question

A flyer on reconnaissance duty spent 5 hours on a mission . He flew out from his base with the wind at the rate of 360mph and returned to his base over the same route, flying against the wind at the rate of mph. How many miles did he fly out before he turned back?

Answer 1

at the rate of mph? the second one?

Answer 2

do you want to learn the material or just get a letter to type in to the computer?

Answer 3

i want to learn

Answer 4

okay; do you know the basic rate equation, rate * time = distance?

Answer 5

please that will be good

Answer 6

yes

Answer 7

I know that rule

Answer 8

ok. also i think you left one of the speeds (in mph) out of the problem?

Answer 9

let me check

Answer 10

A flyer on reconnaissance duty spent 5 hours on a mission . He flew out from his base with the wind at the rate of 360mph and returned to his base over the same route, flying against the wind at the rate of 240mph. How many miles did he fly out before he turned back?

Answer 11

My mistake*

Answer 12

ok, so let's start writing some equations.

Answer 13

yes

Answer 14

so

Answer 15

so first, let's call "t_out" the time he spent flying out, and "t_back" the time he spent flying back.

we know: rate * time = distance, so on the way out,
360 * t_out = d

similarly on the way back
240 * t_back = d

The two d's are the same since he flew the same distance out and back. So we know:

360 * t_out = 240 * t_back.

Does that make sense so far?

Answer 16

yes but t=time right

Answer 17

well I split it into two times, the time he spent flying out and the time he spent coming back. the two times are not the same.

Answer 18

now we have to find t?

Answer 19

there are two different t's in this problem. you use the rate equation separately on the outbound trip and the inbound trip.

Answer 20

360*t=360t
240*t=240t

Answer 21

we need to find how many miles did he fly out before he turned back?

Answer 22

the two t's are different

Answer 23

ok.?

Answer 24

t_out and t_back. they are not the same t.

Answer 25

anyway you know that 360*t_out = 240*t_back, and t_out+tback = 5. now you simultaneously solve those two equations. do you know how to do that?

Answer 26

is this correct?

Answer 27

no, that's not correct; that would be if he flew out for 5 hours and flew back for 5 hours. but that's not what happened. He flew for a *total* of 5 hours.

Answer 28

You don't know how much of that 5 hours was spent going out, and how much coming back. that's what I'm trying to help you solve.

360 * t_out = 240*t_back
t_out + t_back = 5

Can you simultaneously solve those two equations?

Answer 29

I'm not sure... but I think you subtract

Answer 30

subtract what?

Answer 31

you said their both diffrent

Answer 32

yes...? so what are you subtracting?

Answer 33

360-240?

Answer 34

I'm wrong?

Answer 35

no. do you know how to solve systems of equations?

Answer 36

yes

Answer 37

But i need an equation to solve it?

Answer 38

OK so you have to solve the following system of equations:

360 * t_out = 240*t_back
t_out + t_back = 5

can you do that?

Answer 39

No ..

Answer 40

ok so you don't know how to solve systems of equations?

Answer 41

No

Answer 42

well, that's a big topic, but it's something you need to know how to do in order to solve this problem

Answer 43

I know you have to subtract a number to both sides to find the missing variable

Answer 44

you haven't been doing linear equations in other parts of your class?

Answer 45

Well if you simultaneously solve

360*t_out = 240*t_back
t_out+tback = 5
t_out = 5-t_back

360*(5-t_back) = 240 * t_back
t_back = 3
t_out = 2
d = t_out*360 = 720

there you go

Answer 46

just regular equations ;0

Answer 47

A flyer on reconnaissance duty spent 5 hours on a mission . He flew out from his base with the wind at the rate of 360mph and returned to his base over the same route, flying against the wind at the rate of 240mph. How many miles did he fly out before he turned back?
-------------------------------------------
r t = d
Let t be the time to fly to the objective @ 360 mph. To fly back @ 240 mph requires (5-t) hours. The distance out is equal to the distance back.

Solve the following for t:
360 t = 240 (5 - t)
360 t =1200 - 240 t
600 t = 1200
t = 2
360*2 = 720, the miles to fly out.
240(5-2) = 720. the miles to fly back.

Answer 48

Thank you for the medal.

Answer 49

whoa, i gave the answer like an hour ago and got no medal? I don't understand

Answer 50

sorry
i was talking to my tutor hero

Answer 51

np

Answer 52

thanks for your answers

Answer 53

willing to help me in my other question/

Answer 54

yes???????

Answer 55

One last comment.\[r*t = d, t=\frac{d}{r} \]One could solve for d directly using the fact that the sum of the times out and back is five hours.\[\frac{d}{360}+\frac{d}{240}=5 \]However, the arithemetic manipultion is a liitle bit more involved. For example one could multiply each side of the equation by the LCM[360,240]], the lowest common multiple, 720, to obtain after simplification:
5 d = 3600
d = 720 miles

Answer 56

lol thanks I thought you lag problems? at first in till I realized you had a response for me I appreciate your enthusiasm

Answer 57

Thank you. I like to complete answers when possible.

Answer 58

sorry. should have typed: "I like to provide complete answers when possible."

Answer 59

its alright can you go and help me in my other question

Answer 60

yes

Answer 61

Is is the triangle problem?

Answer 62

yes

Answer 63

I'll go there now.

Answer 64

ty