SUM1 tell me wat i am doing WRONG

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The discussion revolves around the calculation of the least distance between two particles, A and B, moving with specified velocities and an initial position vector. The user is struggling to apply the scalar product method correctly, believing they have made an error in their calculations that led to an incorrect time of 0.95 seconds instead of the expected 2.25 seconds. Participants clarify that the position vector of A relative to B should be expressed correctly to find the minimum distance, emphasizing that the motion of A relative to B is a straight line and not necessarily perpendicular to the initial position vector. The conversation highlights the importance of correctly applying vector addition and the scalar product in this context. Understanding the correct formulation is crucial for solving the problem accurately.
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PLZ SUM1 tell me wat i am doing WRONG!

the follwing Q is really annoying me...because i can't work out where i am goin wrong arrgghhhhh

At time t= 0 seconds particles A and B are moving with velocities (-10i-2j)m/s and (-2i-6j)m/s respectively, the postion vector of B relative to A being (-16i+13j)m. Assuming A and B maintain these velocities find the least distance of separation bewteen the 2 particles in the subsequent motion and the value of t for which it occurs.

(PS...we have to use the scalar product method...and yes i can do it with calculus methods...but I am really annoyed i can tget this)

here is my working:

A v B = va-vb = (-8i+4j) - (ie consider situation as "B" being staionary and "A" moving)

B r A (or vector AB)= -16i-13j (ie we are told this)

Vector PB = Vector AB - Vector AP
=(-16i-13j)-t(-8i+4j)
=(-16+8t)i +(-13-4t)j

Given Vector PB will be perpendicular to (-8i+4j) therefore the scalar product of the 2 will = 0

such that : -8(-16+8t)+4(-13-4t) = 0

therfore i get t = 0.95...WHICH IS INCORRECT, the answer is 2.25...can anyone see my ERROE...i really appreciate this!:smile:
 

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ku1005 said:
the follwing Q is really annoying me...because i can't work out where i am goin wrong arrgghhhhh

At time t= 0 seconds particles A and B are moving with velocities (-10i-2j)m/s and (-2i-6j)m/s respectively, the postion vector of B relative to A being (-16i+13j)m. Assuming A and B maintain these velocities find the least distance of separation bewteen the 2 particles in the subsequent motion and the value of t for which it occurs.

(PS...we have to use the scalar product method...and yes i can do it with calculus methods...but I am really annoyed i can tget this)

here is my working:

A v B = va-vb = (-8i+4j) - (ie consider situation as "B" being staionary and "A" moving)

B r A (or vector AB)= -16i-13j (ie we are told this)

Vector PB = Vector AB - Vector AP
=(-16i-13j)-t(-8i+4j)
=(-16+8t)i +(-13-4t)j
Why "- Vector AP"? If AP is the velocity vector of A relative to B and AB the position vector of A relative to B at time t= 0, then the position vector of A relative to B is AB+ APt.

Given Vector PB will be perpendicular to (-8i+4j) therefore the scalar product of the 2 will = 0
I don't follow this. The motion of A relative to B, vector PB, is a straight line. It will not be perpendicular to the initial position vector- that's irrelevant. Instead, minimize the length of vector PB.

such that : -8(-16+8t)+4(-13-4t) = 0

therfore i get t = 0.95...WHICH IS INCORRECT, the answer is 2.25...can anyone see my ERROE...i really appreciate this!:smile:
 
yes i understand that you can simply minimise vector PB...however our topic atm is the scalar product, and it is stipulated that we must use this to answer the question.

“Vector PB = Vector AB - Vector AP” since if you add them as the way u said, it in no way gives vector PB (ie P to B) I agree yes AB+ APt. = however, do u not agree that AP must be in he opposite direction to give the required vector AB….i just skipped this step and inserted the “-“ stright away….

I could hav done ur method and said:


AB+ APt = vector of A relative to B = (-16i+13j)+ -t(-8i+4j)

If u draw it it makes sense since, since isn’t “A relative to B” = vector BA? This is why, again if you draw it, vector BA= (Position vector) A – (position vector) B.


As for the following “I don't follow this. The motion of A relative to B, vector PB, is a straight line. It will not be perpendicular to the initial position vector”

The point at which these 2 particles are closest, given they don’t collide, is the point at which PB is perpendicular to the position vector of t(AvB), ie relative to time hence the line t(-8i+8j), I am not saying it is perpendicular to the initial position vector, I am trying to say it is perpendicular to the position of A relative to B according to time .

Do you follow me now, hence the pic I included, do u not agree that at some point it will be perpendicular and this represents the least distance…….
 
ku1005 said:
yes i understand that you can simply minimise vector PB...however our topic atm is the scalar product, and it is stipulated that we must use this to answer the question.
:bugeye: Are you sure about this?
 
LOL...no not now, caus i know how smart u guys r! its just that that's the way i always hav done finiding that particular vector...so yes i am incorrect, since if you add them as "hallsofivy" stipulates, u obtain the answer, its just that i don't undertsand why it works, tats all...
 
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