Question

hey can someone answer this super hard math problem?

A wagon train that is one mile long advances one mile at a constant rate. During the same time period, the wagon master rides his horse at a constant rate from the front of the wagon train to the rear, and then back to the front. How far did that wagon master ride?

Answer 1

this ur answ34

Answer 2

i love those type of hats what are they called.. i think fedoras

Answer 3

thi sme tryong to figure out math but im not mad about it x

Answer 4

HAHA

Answer 5

ya um idk

Answer 6

can you please stop @ivanapen8

Answer 7

I am trying to get an answer this is due tomorrow

Answer 8

ok sure thing @daniellecrook

Answer 9

got em

Answer 10

tht nice

Answer 11

ok thanks bean

Answer 12

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Answer 13

sweet life

Answer 14

this tasted good

Answer 15

http://answers.yahoo.com/question/index?qid=20080701115439AAyLlXW its ok i dont need ur answers

Answer 16

this was the soup

Answer 17

thats our math teacher IM LAUGHING

Answer 18

that looks really yummy mm make sure u dont cut yourself on that plate!

Answer 19

HA

Answer 20

matt just left the house

Answer 21

these thinkgs arent good

Answer 22

ok???????????

Answer 23

tbh its been showing this for the last hour

Answer 24

tht a no brainer

Answer 25

just do it

Answer 26

oldrin type

Answer 27

send it

Answer 28

wow r u kidding

Answer 29

HAHAHHA

Answer 30

oldrin is still typing the mathematician is hard at work

Answer 31

Let's say the wagon masters absolute speed is \(r_m\), and so his effective (relative) speed towards the rear of the train is given by \(r_m+r_w\) (the train is approaching him). His relative speed towards the front of the train, however, is only \(r_m-r_w\) (the train is moving away from him).
It should make sense that going towards the rear, our biker effectively travels \(1\) mile at a rate of \(r_m+r_w\) and hence it takes him \(\dfrac1{r_m+r_w}\). On the way back, however, he effectively travels \(1\) mile at a rate of \(r_m+r_w\) and it takes him \(\dfrac1{r_m-r_w}\).

Consider, then, that total amount of time taken must be merely the time it took the wagon itself to travel that far i.e. \(\frac1{r_w}\) hence:$$\frac1{r_m+r_w}+\frac1{r_m-r_w}=\frac1{r_w}\\\frac{r_m-r_w+r_m+r_w}{r_m^2-r_w^2}=\frac1{r_w}\\2r_mr_w=r_m^2-r_w^2\\r_m^2-2r_mr_w-r_w^2=0\\r_m^2-2r_mr_w+r_w^2=2r_w^2\\(r_m-r_w)^2=2r_w^2\\r_m-r_w=r_w\sqrt2\\r_m=(1+\sqrt2)r_w$$Thus the master travels at the absolute rate of \((1+\sqrt2)r_w\) and so in the same time span (i.e. \(1/r_w\)) travels \((1+\sqrt2)r_w\times1/r_w=1+\sqrt2\) miles.

Answer 32

typo: he effectively travels at a rate of \(r_m\color{red}-r_w\) on the way back to the head of the wagon train

Answer 33

ty so much