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CATCHING SMUGGLERS physics question

This is the question: https://isaacphysics.org/questions/smuggler_catching_spiral?board=5dc86f7b-470c-4dab-91da-4703456f22bb&stage=all

I have tried integrating k*v*sin(theta) dt, as the video suggests, but I think I should use some integration factor, as theta also changes with time.
Original post by javier garcía
This is the question: https://isaacphysics.org/questions/smuggler_catching_spiral?board=5dc86f7b-470c-4dab-91da-4703456f22bb&stage=all
I have tried integrating k*v*sin(theta) dt, as the video suggests, but I think I should use some integration factor, as theta also changes with time.

Not got time to spend on it, but suggest having a Google on "pursuit curves".
You should hit a close match on the first page.
(edited 9 months ago)
Reply 2
Original post by Javier García
This is the question: https://isaacphysics.org/questions/smuggler_catching_spiral?board=5dc86f7b-470c-4dab-91da-4703456f22bb&stage=all
I have tried integrating k*v*sin(theta) dt, as the video suggests, but I think I should use some integration factor, as theta also changes with time.

You can form 2 integrals, one for the net distance travelled towards the smuggler boat, and one for the distance travelled in the y-direction (perp to the shore).

These integrals both involve sin(theta) and at the point of interception they both equal a. By multiplying and adding (similar to simultaneous equations) you should be able to eliminate the sin theta term and you're left with a quadratic to solve.
Original post by DFranklin
You can form 2 integrals, one for the net distance travelled towards the smuggler boat, and one for the distance travelled in the y-direction (perp to the shore).
These integrals both involve sin(theta) and at the point of interception they both equal a. By multiplying and adding (similar to simultaneous equations) you should be able to eliminate the sin theta term and you're left with a quadratic to solve.

Thanks to both of you guys!
Sorry to jump in but If I could just ask a couple of questions...

I got the answer to this problem but I dont really understand why it works.
I watched the video https://www.youtube.com/watch?v=_x0hSuZi8Js
and I get how the relative parallel velocity is kv - vsin(theta)
but he says "only the parallel component changes the distance of the two boats" which made sense but after thinking about it I dont really get it?

In this diagram I split the velocity into parallel and perpendicular components
working out.png

and it should imply that v⊥ should move the smuggler ship from the black dot to the red dot, so increasing its distance from the coastguard ship but it doesnt have any effect on the distance?
My question basically is how can we ignore v⊥?

I also dont get the first integral he set up (I understand the other 2)
Int v_rel dt = a
I get how Int kvcos(theta) dt = a
but that doesnt seem to be important - how does the relative velocity (that too one which neglects v⊥) have a distance equal to a?

Any explanations greatly appreciated.
Reply 5
Original post by mosaurlodon
Sorry to jump in but If I could just ask a couple of questions...
I got the answer to this problem but I dont really understand why it works.
I watched the video https://www.youtube.com/watch?v=_x0hSuZi8Js
and I get how the relative parallel velocity is kv - vsin(theta)
but he says "only the parallel component changes the distance of the two boats" which made sense but after thinking about it I dont really get it?
In this diagram I split the velocity into parallel and perpendicular componentsworking out.pngand it should imply that v⊥ should move the smuggler ship from the black dot to the red dot, so increasing its distance from the coastguard ship but it doesnt have any effect on the distance?
My question basically is how can we ignore v⊥?
I also dont get the first integral he set up (I understand the other 2)
Int v_rel dt = a
I get how Int kvcos(theta) dt = a
but that doesnt seem to be important - how does the relative velocity (that too one which neglects v⊥) have a distance equal to a?
Any explanations greatly appreciated.

Suppose S (smugglers) have position r = (x, y) relative to C (coast guard). Then the relative speed is dr/dt = (dx/dt, dy/dt).
If D is the square of the distance from S to C then D = x^2 + y^2. Differentiate w.r.t and you get dD/ dt = 2x dx/dt + 2 y dy/dt, which is 2r . dr/dt.
So if you decompose dr/dt into components P parallel and N normal to r, then the only component affecting dD/dt (and therefore the (rate of change of the) distance from S to C) is the parallel component (since N.r = 0).

The "P component" of the velocity has size kv cos theta. This is the rate at which C is approaching S. And so kvcosθdt\int kv \cos \theta \,dt is the total amount by which C has closed in on V. But initially C is distance aa from S, and when C catches S, the distance is 0. So C has closed in by aa meters, and this must be the value of the integral.

Edit: (I haven't actually watched the video).
(edited 8 months ago)
I tried to follow what you said closely but Im not sure how to get N as a vector form to make N dot r = 0
Is this working out right so far?
working out.png


so basically the video sets up the parallel relative velocity v_relative = kv - vsin(θ)
and then the integral
v_relative dt = a
which comes from the fact that the horizontal distance travelled = a so
kvcos(θ) dt = a
right?
Reply 7
Original post by mosaurlodon
I tried to follow what you said closely but Im not sure how to get N as a vector form to make N dot r = 0
Is this working out right so far?
working out.png
so basically the video sets up the parallel relative velocity v_relative = kv - vsin(θ)
and then the integral
v_relative dt = a
which comes from the fact that the horizontal distance travelled = a so
kvcos(θ) dt = a
right?

You don't need to "get N in vector from". By definition, it's the same as your v⊥. The point I'm trying to make is that if you write dr/dt = P + N , where those are the components parallel and perpendicular to R, then d/dt D only depends on P, not N, because r.N = 0.

And that is why "only the parallel component changes the distance".

If it helps, it's also very similar to why when you have an object moving in a circle, the radial acceleration only causes the direction of motion to change, it doesn't change the speed.
Ohhh I see what you mean now.
so dD/dt = 2r.dr/dt = 2(r.N+r.P)=2r.P

and I think I also get why the integral of (the relative velocity) P has that specific value but let me just reiterate to solidify the points (and to see if I made any mistakes)
Say v_relative is the parallel velocity of C relative to S
we've already established that only the relative parallel velocity can change the distance between the boats.
Initially they were distance A apart, and then a distance 0 apart.
meaning that the relative parallel velocity had to have closed a distance of A.
thus
v_relative dt = A

Originally I thought that this integral represented the "curved distance" travelled by S, ie the red line
working out.png
but then realised this was the integral
kv dt
and the the integral v_relative dt actually represents the distance closed which = A.

Thank you so much

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