Mechanics, i's and j's

  • Thread starter NinaL
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I came across this question in a past paper, and am absolutely stuck.

A coastguard station 0 monitors the movements of ships in a channel. At noon, the station's radar records two ships moving with constant speed. Ship A is at the point with position vector (-5i + 10j) km relative to 0 and has velocity (2i + 2j) km/h. Ship B is at the point with position vector (3i + 4j) km and has velocity (-2i + 5j) km/h.

(a) Given that the two ships maintain these velocities, show that they collide.



Is there anyone who can help? Even if it's just tips on where to start etc?
Thanks in advance!
 

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  • #2
dextercioby
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NinaL said:
I came across this question in a past paper, and am absolutely stuck.

A coastguard station 0 monitors the movements of ships in a channel. At noon, the station's radar records two ships moving with constant speed. Ship A is at the point with position vector (-5i + 10j) km relative to 0 and has velocity (2i + 2j) km/h. Ship B is at the point with position vector (3i + 4j) km and has velocity (-2i + 5j) km/h.

(a) Given that the two ships maintain these velocities, show that they collide.



Is there anyone who can help? Even if it's just tips on where to start etc?
Thanks in advance!
The problem is not difficult.
HINTS
1.[tex] \vec{r}(t)=\vec{r}(0)+\vec{v}t [/tex] (1)
2.The condition that the two ships collide reads
[tex] \vec{r}_{1}(t)=\vec{r}_{2} (t) [/tex](2)
3.Combining (1) with (2) will bring u an algebraic system of 2 equations with one unknown.For the crash to happen,they need to have the same solution.


Daniel.
 
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Wow, I feel so stupid!
Sorry, but can you explain the symbols that you've used? :)
Sorry!


Thank you for the help though!
 
  • #4
dextercioby
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Look at the bright side...It's better to feel stupid,than to be stupid...
[itex] \vec{r} [/itex] is the position vector of a body.[itex] \vec{r}_{1} [/itex] is the position vector for the first ship.[itex] \vec{r}_{2} [/itex] is the position vector for the second ship.In order to collide,the 2 ships must have the same position,which means the same position vector.So that explains the second equation.
The first equation gives the connection between the position vector of a body,the velocity vector and 'the time'.
U'll need to write two equations similar to the first:eek:ne for the first ship,one for the second.And then,simply ask that that the position vectors of the two vectors coincide.

Daniel.
 
  • #5
BobG
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dextercioby said:
The problem is not difficult.
HINTS
1.[tex] \vec{r}(t)=\vec{r}(0)+\vec{v}t [/tex] (1)
2.The condition that the two ships collide reads
[tex] \vec{r}_{1}(t)=\vec{r}_{2} (t) [/tex](2)
3.Combining (1) with (2) will bring u an algebraic system of 2 equations with one unknown.For the crash to happen,they need to have the same solution.


Daniel.
Daniel's first equation is just your position equation for vectors(s_f=s_i+vt+1/2at^2), except there's no acceleration in this case.

Your initial position is s_i and your velocity is given in vector format. So, you'd have:

[tex]\vec{r}(t)=(-5_i + 10 _j) + (2_i + 2_j)t [/tex]
Note that the velocity vector could also be written: [tex](2t_i + 2t_j)[/tex]
Per vector addition, your final position vector is:
[tex]\vec{r}(t) = (-5+2t)_i + (10+2t)_j[/tex]

You do the same for the second boat.

Daniel's 2nd equation says that at some given instant, both vector sums are equal. In other words, subtracting one vector from the other will give you a null vector (0_i + 0_j). By combining terms you should be able to figure out one value for t that will zero out both your i and j terms. If you can, they collide. If they can't, they won't collide.
 

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