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Go ahead and ask please.

 

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.)

 

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.

 

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.

 

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.

 

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.

 

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.)

 

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