Question

a chicken starts running eastwar from a point 30 meters due west of a telephone pole. The chicken runs at 4.1 meters per second. At the same instant, a gopher starts from a point 40 meters north of the pole, heading south at 3.8 meters per second. After how much time will the chicken and gopher first be 20 meters apart?

Answer 1

Why is the chicken running?

Answer 2

Draw picture.

Answer 3

I did

Answer 4

stil dont get it

Answer 5

The chicken's position is given by: \[
f(t)=4.1t
\]The gopher's position is given by: \[
g(t)=40-3.8t
\]

Answer 6

The distance between them is \[
d(t) = \sqrt{[f(t)]^2+[g(t)]^2}
\]

Answer 7

the chicken is 30 meters from the pole

Answer 8

You need to solve for \(t\) where: \[
20=\sqrt{(30+4.1t)^2+(40-2.8t)^2}
\]

Answer 9

Oh wait...

Answer 10

You need to solve for \(t\) where: \[
20=\sqrt{(30-4.1t)^2+(40-2.8t)^2}
\]

Answer 11

http://www.wolframalpha.com/input/?i=20%3D%5Csqrt%7B%2830-4.1t%29%5E2%2B%2840-2.8t%29%5E2%7D

Answer 12

You need to find the lowest \(t\) where \(t>0\)

Answer 13

I get \(t\approx 7.15\)

Answer 14

i know the answer, but dont know how to do it. And by the way, it is 5.6 seconds

Answer 15

Now I'm getting \(t=5.6\) using: \[
20=\sqrt{(30-4.1t)^2+(40-3.8t)^2}
\]

Answer 16

\(\color{blue}{\text{Originally Posted by}}\) @wio
The distance between them is \[
d(t) = \sqrt{[f(t)]^2+[g(t)]^2}
\]
\(\color{blue}{\text{End of Quote}}\)
This is the essential concept to use.

Answer 17

You just need to model the position of each creature.

Answer 18

so you endded up getting 5.6 seconds with the formula//
?

Answer 19

Yes.

Answer 20

thank you very much! I raelly needed it

Answer 21

Okay, but I don't know why chicken is running.

Answer 22

the world may never know