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Hello Everyone:
I know it is quite late, but i was googling for help with a physics question and i stumbled upon this site...
I know it is quite late, but i was googling for help with a physics question and i stumbled upon this site...
thanks..phreak said:Go ahead and ask please.
phreak said: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?
phreak said:No. It means they move east, then south.
yes..phreak said: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?)
which one...phreak said:@[email protected] Well, that question was quite confusing, but yes... I think so.
phreak said: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.
phreak said: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.)
thanks for your help.phreak said:Alright. Then all of your questions should be answered (temporarily, at least). At that, I'm going to bed. ^_^