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 November 12th, 2017, 01:41 AM #1 Newbie   Joined: Nov 2017 From: Serbia Posts: 2 Thanks: 0 Sea ships and their space-time equations On different days, two sea ships and their space-time equations (coordinates in km and hours) were observed: Find out whether the boats have different velocities or different courses. My question: Since the coordinates are time and space, how do I plot these on a graph? Can I do the following: For example for A, can I take 12 as x1 space coordinate and 8 as y1 time coordinate of one point. Then, take -4 and 18 for the other point so that could be the first vector? And how can I calculate the velocity? Thank you. November 12th, 2017, 04:22 AM   #2
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 Originally Posted by Tarata12 On different days, two sea ships and their space-time equations (coordinates in km and hours) were observed: Find out whether the boats have different velocities or different courses. My question: Since the coordinates are time and space, how do I plot these on a graph? Can I do the following:
Since the independent variable is time, one dimensional, and the dependent variable, is position, two dimensional, in order to graph this you would need a three dimensional graph. Not impossible but awkward to graph, especially if you are doing it on two dimensional paper!

Quote:
 For example for A, can I take 12 as x1 space coordinate and 8 as y1 time coordinate of one point. Then, take -4 and 18 for the other point so that could be the first vector? And how can I calculate the velocity? Thank you.
I have no idea where you got "8" and "18" as time coordinates! When t= 0, the A position is (12, -4). When t= 1, the A position is (24, -8 ). If t= 8 (just because you mention it) the A position is (12+ 8(12), -4+ 8(-4))= (108, -36). But that really has nothing to do with this problem.

In any case, rather than using three dimensions to graph A's position, I would say that since x= 12+ 12t, t= (x- 12)/12 and then y= -4- 4t= -4- 4(x-12)/12= -4- (x- 12)/3= -x/3. The graph is y= -x/3, a straight line (you could then, mark points on that line with their t value. For example, if t= 0, x= 12, y= -4 so the point (12, -4), which lies on y= -x/3, corresponds to t= 0.)

Similarly Since, for B, x= 10+10t, t= (x- 10)/10. Then y= -2+ 25t= -2+ 25(x- 10)/10= -2+ 2.5x- 25= 2.5x- 27.

Notice that those lines have different slopes, -1/3 and 2.5, so they are definitely not the same course. As far as the velocities are concerned they are just the vectors (12, -4) and (10, 25). Of course, -4/12= -1/3 and 25/10= 2.5. November 12th, 2017, 05:25 AM #3 Newbie   Joined: Nov 2017 From: Serbia Posts: 2 Thanks: 0 Ah, I see. I made a mistake in the equation, it's actually $x = [12 -4] +t*[8, 18]$ that's why you couldn't find where it comes from Anyway, it was not correct. Now I understand, I solved the problem Thank you very much! November 12th, 2017, 12:56 PM #4 Global Moderator   Joined: May 2007 Posts: 6,704 Thanks: 670 the speeds are different - 26.9 for B and 19.7 for A. B's speed components have a ratio of 2.5, while A's have a ratio of 2.25, so they are not on the same course. Tags equations, sea, ships, space-time, spacetime, vectors Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post lemgruber Physics 9 August 17th, 2015 12:12 PM HannahJane1993 Differential Equations 2 April 22nd, 2015 02:09 AM Dacu Math Events 16 July 10th, 2013 11:13 AM brambram Applied Math 0 October 19th, 2012 07:57 AM hml Real Analysis 2 October 21st, 2009 09:12 AM

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