The closest approach of two ships

[gnits]

Homework Statement

To find the closest approach of two ships

Relevant Equations

d=st

Could I please ask for help with the following question:

A destroyer moving on a breaing of 30 degrees at 50km/h observes at noon a cuiser traveling due north at 20km/h. If the destroyer overtakes the cruiser one hour later find the distance and bearing of the cruiser from the destroyer at noon.

(answer is given as 34.2km , 47 degrees)

(I am assuming that "overtaking" means the point a which the distance between the ships is minimal)

So, here's my attempt:

Let i be the unit vector in the direction of North and j be the unit vector in the direction of East.
Let position vector of detroyer = $r_d$
Let position vector of cruiser = $r_c$
Let position vector of destroyer relative to cruiser = $r_{dc}$
Let the origin of my coordinate system be the position of the destroyer at noon

Let position of cruiser relative to detroyer at noon be (h,d)

Then:

$r_d=50\,sin(30)t\,i + 50\,cos(30)t\,j\,=25ti+25\sqrt{3}tj$

and

$r_c=hi+(d+20t)j$

So, as we are told that the overtaking happens at t = 1, then we know that overtaking happens at:

$r_d=25i+25\sqrt{3}j$ and $r_c=hi+(d+20)j$

Also, we know that at any time t:

$r_{dc} = (25t-h)i+(25\sqrt{3}t-20t-d)j$

And so the square of the distance between the ships at any time is given by:

$D^2=625t^2-50ht+h^2+d^2-50\sqrt{3}dt+40dt=1000\sqrt{3}t^2+2275t^2$

So distance will be a minimum when the differential of this with respect to time is 0, i.e. when:

$(4-5\sqrt(3))d-5h-(200\sqrt{3}+580)t=0$

And we are told that this occurs when t = 1 and so we have:

$\frac{(4-5\sqrt{3})d-5h}{200\sqrt{3}-580}=1$

(If I plug in the answer here of $h = 34.2\,sin(47)$ and d = $34.2\,cos(47)$ then this gives 1, which is good)

Not sure how to proceed from here to find h and d.

If I could somehow reason that the overtaking will happen when then destoyer is due south of the cruiser then I could know that here h = 25 and so solve the above equation to get d = 23.3 which would lead to the correct answer.

Thanks for any help,
Mitch.

 

[PeroK]

Homework Statement:: To find the closest approach of two ships
Relevant Equations:: d=st

Could I please ask for help with the following question:

A destroyer moving on a breaing of 30 degrees at 50km/h observes at noon a cuiser traveling due north at 20km/h. If the destroyer overtakes the cruiser one hour later find the distance and bearing of the cruiser from the destroyer at noon.

(answer is given as 34.2km , 47 degrees)

(I am assuming that "overtaking" means the point a which the distance between the ships is minimal)

So, here's my attempt:

Let i be the unit vector in the direction of North and j be the unit vector in the direction of East.
Let position vector of detroyer = $r_d$
Let position vector of cruiser = $r_c$
Let position vector of destroyer relative to cruiser = $r_{dc}$
Let the origin of my coordinate system be the position of the destroyer at noon

Let position of cruiser relative to detroyer at noon be (h,d)

Then:

$r_d=50\,sin(30)t\,i + 50\,cos(30)t\,j\,=25ti+25\sqrt{3}tj$

and

$r_c=hi+(d+20t)j$

So, as we are told that the overtaking happens at t = 1, then we know that overtaking happens at:

$r_d=25i+25\sqrt{3}j$ and $r_c=hi+(d+20)j$

I would assume that by overtaking we mean the destroyer intercepts the cruiser. In which case, you nearly have the answer here.

 

[gnits]

Thanks for your reply PeroK, The question is indeed from a section on interception and within that context your assumption makes good sense, and indeed I believe it to be the intention of the question. I think the use of the word "overtakes" is confusing, it should be "intercepts". Thanks for your help, Mitch.

 

[jbriggs444]

A destroyer moving on a bearing heading of 30 degrees at 50km/h observes at noon a cuiser traveling due north at 20km/h. If the destroyer overtakes the cruiser one hour later find the distance and bearing of the cruiser from the destroyer at noon.

A heading is where a ship is pointing. A bearing is the direction to something. [A "course" is slightly different from a heading -- it take current into account, but there are no currents here]

The questioner needs to brush up on navigational terminology.

 

[PeroK]

A heading is where a ship is pointing. A bearing is the direction to something. [A "course" is slightly different from a heading -- it take current into account, but there are no currents here]

The questioner needs to brush up on navigational terminology.

I listened to the shipping forecasts on Radio 4 for years before I learned the difference between "veering" and "backing".