M1 vectors...

How are you supposed to do this? The answer says to use s = ut + 0.5at^2 for A. but then what is s? Is this the displacement vector of s? How do you know that this is where the particles collide?

And also how would you find the position vector of B's starting point?

Why do you want that

okay thanks. its part of the questio

Well they have the same position vector

So you do the same and figure out what has been added on


Wow ... negged for helping ... great

Thanks. I didn't neg you!

No, I know


DId you solve the question?

Yes I have another q though. If you have 2 particles A and B and want to find when B is due east of A, why is it that the i coefficients are equal? And why is it that the j coefficients are equal if you want to find when B is due north of A? (If you are given their position vectors) I can't get my head around this.

And how do you find velocity of the particles if you are only given position vectors of them?

Finally, if 2 vector quantities C and D collide (e.g. if C intercepts D at a certain time) how do you find the resultant vector velocities involved? is it that the vector CD = 0, or DC = 0?
And to calculate CD, is it position vector D - position vector C?


Suppose it would help if I gave a question for context of the intercepting thing:

at 0900 a boat is travelling at constant speed, with position vector -2i - 4j km relative to the origin.
0940 S is at 4i - 6j.

Find an expression for s in terms of t

At 1100 a motorboat leaves O and travels with constant velocity (pi + qj) kmh^-1. If it intercepts the first boat at 1130, calculate p and q.

That depends on the definition of i and j



Velocity is (change in position)/time

Moving from 3i+5j to 7i+15j in 2 seconds
Change in position = 4i+10j
Divide by time = 2i+5j = velocity



Boat has velocity = (6i-2j)/{4/6} = 9i-3j
r = (-2i-4j) + (9i-3j)t
So at 11 r = (-2i-4j) + (9i-3j)2 = 16i-10j

SO

Starting at 1100
B = (6i-10j) + (9i-3j)t
M = (0i+0j) + (pi+qj)t

They meet at t = 0.5

i gives 6 + 4.5 = 0 + 0.5p

etc