# Kinematic One-D

1. Jan 17, 2012

### saac

1. The problem statement, all variables and given/known data

A women and her dog are out for a morning run to the river, which is located 4km away. The women runs at 2.5m/s in a straight line. The dog is unleashed and runs back and forth at 4.5m/s between his owner and the river, until the women reaches the river. What is the total distance run by the dog?

2. Relevant equations
Average speed=d/t
not sure what else.

3. The attempt at a solution
I figured that you have to figure out the total time it takes the dog to make it to the river. Which is 888.9s(4000m/4.5m/s). Then calculate where the women is when the dog reaches the river. So 2.5m/s*888.9=2222.25m.... I do not know how to calculate the dogs return to the women while she is still moving at 2.5m/s. Help please. Got a week to learn kinematics

2. Jan 17, 2012

### Simon Bridge

You could do it that way - and solve for the infinite sum ... or you can realize that the distance the dog runs is equal to the speed of the dog multiplied by the time the dog runs for :)

3. Jan 18, 2012

### saac

So would the distance the dog has to travel 4000m-2222.25(where women is when dog reaches river)

So d=4.5m/s*888.9s? that brings me back to the total distance. Or would d=4.5m/s*395.1s? the 395.1s obtained by taking 4000m-2222.25m=1777.75m then t=1777.75m/4.5m/s=395.1s? Am I on the right track...

Last edited: Jan 18, 2012
4. Jan 18, 2012

### Simon Bridge

Nope - you are doing the opposite of what I said: the path the dog followed doesn't matter.
If the dog runs at speed v for time t, then the total distance the dog runs is d=vt
You are given v. You need t. Find t.

You'll kick yourself when you see it.
Hint: the other things you know are the speed of the woman and the distance to the river.

Last edited: Jan 18, 2012
5. Jan 18, 2012

### saac

I am not sure how to word this but here's what i got...1777.75(distance the women is from the river once the has arrived)/2.5m/s(her rate)=711.1s(time it will take for her to arrive). Then i used that time to solve for the dogs D- d=4.5m/s*711.1s= 3199.95m. If i take this distance and add to it the distance the dog has already traveled(4000m) i end up with 7199.95 for the total distance the dog has traveled. Closer?

been kicking...

6. Jan 18, 2012

### Simon Bridge

ARGH! Where women is when dog reaches river DOES NOT MATTER!!!!!

I cannot think of a clearer way to put it without doing your homework for you.
Use d=(4.5)t ... you need to find t .... focus on that.
Forget about where the woman is at special times.

You have to find what t is from the other information supplied.
You are given the speed of the woman and the distance to the river ... if only there was a way to find the time from the speed and the distance?!

Last edited: Jan 18, 2012
7. Jan 18, 2012

### saac

so find the time for both? t=4000/2.5= 1600 & t=4000/4.5=888.9 then the difference between the two, 1600-888.9=711.1. I'll use this time to find the dogs distance? 711.1*4.5...and the rest is the same a my last post.

my apologies for my complex "reasoning". gotta work out the bad before you usher in the good, right? lol

8. Jan 18, 2012

### Simon Bridge

Why do you need the difference between the two times?
(This is like watching a pro golfer continue to miss an easy putt.)

Remember - the path of the dog does not make any difference.
I could say: the woman walks to the river - the dog runs to a distant tree and back to the woman.
What is the total distance the dog runs?

You need to give up these preconceptions young padawan :)
Put pen and paper away and sit back and clear your mind - you are too stuck on needing a difference in something - times, distances, whatever. When you are calm, clear these things will become.

On the time, you will focus... on the time...
The dog has a limited amount of time to run around in.
(pause - breathe - your brain will want to go joyously about it: stop - calm - focus.)

The dog starts running when? ... when the woman begins her journey.
The dog finishes running when?... when the woman ends her journey.
So - how much time is that?

This is actually an important conceptual lesson - absolutely everyone starts out stuck in the same sort of mindset you are in.
The point of the exercise is to create a paradigm shift ... which is hard and painful but worth it the shift is - yes.

Last edited: Jan 18, 2012
9. Jan 18, 2012

### saac

Ah damn slightly shocked i didn't notice that. so dog is moving for 1600s at 4.5m/s. d=1600s*4.5m/s=7200m

Certainly is a time consuming and painful shift. Thanks though!

10. Jan 18, 2012

### Simon Bridge

Ah but now you can go find your classmates and watch them struggle through the same thing! Srsly - nobody sees it. The question is asked in a way that is designed to misdirect your attention to things that don't matter. You will also meet the effect when examining pseudoscience claims - particularly in the free-energy crowd. However, it is usually the way with real life problems - they don't come packaged, so you have to look through them to find out what is important. It is an art.

Now you are going to have trouble not seeing it.

Notice how you kept almost getting it right and just missing?
The closest you got was 7199.95m - due to rounding errors.
If you have a look at that you'll see how you got so close.

The main lesson is about overcomplicating the situation.

11. Jan 19, 2012

### saac

Well you are right- It is an art. I'll keep trying, although the struggle will never end...btw you must have oodles of patience to deal with all the noob questions, or more specifically, the noobs doing the questions.

12. Jan 19, 2012

### Simon Bridge

There are no stupid questions - they are clues to understanding.

I've never had to do this one in text before - f2f I've had students actually sitting cross-legged and going "om" to reset their minds. When I'm right there I can do things like point and wave my arms so I was at a bit of a loss.

From here you should feel better about puzzles.
Go find some to do.
Have fun.