Havin' problem in vector direction

[kanki]

What i have now is the initial relative position of A from B, which is 5j
initial velocity of A is 40i + 30j, the direction is changed so that it can overtake B.
Velocity of B is 12i + 16j.
I can find the speed of A, which is 50.
I tried to let the new velocity of A be 50(ai + bj), then here's the problem:
I tried to use
$$\\vec{r}_A = \\vec{r}_B$$
r_A=r_B for which r represents the displacement.
but there are 3 unknowns and I'm stuck there...
*I'm having problem with the latex code... hope u can understand

 

[mathmike]

so what are you trying tofind

 

[kanki]

I'm trying to find the new direction of velocity of A

 

[mathmike]

so do this

a is at the orgin

then b is at 0,5

if the vel of b is 12i + 16j

where would that put it in the next unit of time

 

[kanki]

I'm sorry, the s should be located north of b,
so b is the origin, and a at (0,5).
If vel of b is 12i + 16j, then in next unit of time,
it'll reach the point (12,16).
What can it help me?
I still dun understand.

 

[mathmike]

ok now what would be the vector be to get a to that point?

 

[kanki]

12i + 11j?
u mean for A?
still dun understand.

 

[mathmike]

if you are going from 5j to 12i + 16j then the vector be 12i + 16j

 

[kanki]

What can i get from vector 12i + 11j, what to do next?
Sorry I'm too dumb... i need more explanation if possible...

 

[kanki]

helo?
no one can help me?

 

[verty]

What do you want to know? I don't understand this notion of 'to overtake B'. Do they mean to meet B (as in, calculating the angle to fire a bullet to hit a moving target)?

 

[verty]

Well I'm guessing you want to know the new direction of A, such that A will meet B, like a bullet hitting a moving target.

I drew an obtuse triangle with sides {5,2x,5x}, then used the cos-rule (one angle is known) to get an equation in x, which I solved. Knowing x, one knows the lengths and finding the angle is just a small step away.