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[SpicyQ]

Hello Everyone:

I know it is quite late, but i was googling for help with a physics question and i stumbled upon this site...

 

[phreak]

Go ahead and ask please.

 

[SpicyQ]

Go ahead and ask please.

thanks..

its quite simple (im just stuck)

it says the dashed line is the path of a walker wandering through the woods, finally arriving at the co-ord. (x,y) = (300, 400) meters

what is the displacement from the origin

info:

the graph is a m-m graph and the walker walks in an S-pattern from the origin. (i can't put the graph up, but that what it looks like)

since it is not a straight line and no time involved, how to i get the magnitude?

locations 1 is (0,0)
location 2 is (300, 400)


i'll put up what i did in a min.. i just wanted to make sure i was in the right track..

thanks ( sorry for the lenght)

 

[SpicyQ]

i got 500 for my magnitude and 53.1 for my direction

and my v-ave = 0.5m/s

 

[phreak]

This is involving a vector, I'm assuming?

With vectors, the path a particle journeys does not matter, so whether it is a straight line or a curvy line, the only things that matter are the end points, that is, the initial location and the final location.

So, all you must do is look at the end points and find the magnitude and direction. That is, sqrt(300^2+400^2) is the magnitude, right? And from that, finding the direction should be quite simple if you know your trigonometry.

Or was this the answer you were looking for?

(EDIT: Yes, your first answer is correct, although I don't know where you generated your second answer, as because there is no time component (or any other component), an average velocity is impossible to come by.)

 

[SpicyQ]

This is involving a vector, I'm assuming?

With vectors, the path a particle journeys does not matter, so whether it is a straight line or a curvy line, the only things that matter are the end points, that is, the initial location and the final location.

So, all you must do is look at the end points and find the magnitude and direction. That is, sqrt(300^2+400^2) is the magnitude, right? And from that, finding the direction should be quite simple if you know your trigonometry.

Or was this the answer you were looking for?

thanks a lot..
thats was how i did it..
one more question as far as stating my direction.

when one says they are located "south of east"...does that mean that from their current position, they move south, and then move east?

 

[phreak]

No. It means they move east, then south. For example, 15 degrees south of east would be 115 degrees from 0 degrees North, going clockwise.

 

[SpicyQ]

No. It means they move east, then south.

would saying 53 degrees south of east be a good representation?


(thats what i got when i did my calculations)

 

[phreak]

I did not see any unit of direction on your problem. Therefore, I would say no. You would most likely say something such as "53.1 degrees along the x-axis" for a final answer; that is, that would be the correct answer if the problem was stated correctly (the final coordinates are located in the first quadrant, right?)

Also, although I don't encourage it, most teachers will just accept the final answer as "53.1 degrees" without anything following it.

 

[SpicyQ]

I did not see any unit of direction on your problem. Therefore, I would say no. You would most likely say something such as "53.1 degrees along the x-axis" for a final answer; that is, that would be the correct answer if the problem was stated correctly (the final coordinates are located in the first quadrant, right?)

yes..
the question asked to find mag + direction

i used the sqrt eqn to find the magnitude,
i found the direction in the x-direction by using Dx/D =cos (teta)

i did not find direction in the y direction

i think i should have drawm out the vectors and found the resultant and then the direction of the resultant would be my answer?

 

[SpicyQ]

since its a straight line (when i connect the starting point to the finish point) i have to find direction in x & y axis?

 

[phreak]

As for your first question, @_@ Well, that question was quite confusing, but yes... I think so.

As for your second question, no, you shouldn't have to... particularly because the answer is a given. The x-component is always exactly on the x-axis and the y-component is always exactly on the y-axis.

 

[SpicyQ]

@_@ Well, that question was quite confusing, but yes... I think so.

which one...

 

[phreak]

I've gotten to where I'm confusing myself. You mentioned the correct answer above. That is, "500m at 53.1 degrees along the x-axis." Use that one.

 

[SpicyQ]

I've gotten to where I'm confusing myself. You mentioned the correct answer above. That is, "500m at 53.1 degrees along the x-axis." Use that one.

thanks..

and my average velocity would be 0.5m/s

how would i get the magnitude and direction of the average velocity?

 

[phreak]

How did you compute average velocity? There is no component of time to work with in your initial problem, therefore there cannot be an average velocity. If the time you are using is t = 1000s, then 0.5 m/s would be the correct average velocity. Otherwise, no.

However, I'll answer your second question anyway, just in case I'm missing anything. The direction of the average velocity will be the same as that of your other velocity. The magnitude would be 0.5 m/s.

(And as for my above paragraph of text, I meant that your final answer should be 500 m/s, not just m.)

 

[SpicyQ]

How did you compute average velocity? There is no component of time to work with in your initial problem, therefore there cannot be an average velocity. If the time you are using is t = 1000s, then 0.5 m/s would be the correct average velocity. Otherwise, no.

However, I'll answer your second question anyway, just in case I'm missing anything. The direction of the average velocity will be the same as that of your other velocity. The magnitude would be 0.5 m/s.

(And as for my above paragraph of text, I meant that your final answer should be 500 m/s, not just m.)

cool

it's a 2-part problem

part a did not have component of time

part be asked to assume that the walk took 1000 seconds

 

[phreak]

Alright. Then all of your questions should be answered (temporarily, at least). At that, I'm going to bed. ^_^

 

[SpicyQ]

Alright. Then all of your questions should be answered (temporarily, at least). At that, I'm going to bed. ^_^

thanks for your help.
good night :zzz: