Sea ships and their space-time equations

Nov 2017
2
0
Serbia
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.
 

Country Boy

Math Team
Jan 2015
3,261
899
Alabama
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!

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.
 
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Nov 2017
2
0
Serbia
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!
 

mathman

Forum Staff
May 2007
6,932
774
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.