- #1
gnits
- 137
- 46
- 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.
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.