# Relative velocity question

1.Two particles A and B have velocities 3i and vj respectively (in m/s).

(a) Find the position of B relative to A for all t given that r (b relative to a)
(t=0) = -9i+6j (in metres).

(b) Find the value of v such that A and B collide.

(c) If v=1 m/s, find the time and distance when A and B are closest together.

3. Well, firstly i drew it out. A= 3i+0j , B=0i+vj. Then I integrated A and B and let it equal to -9i+6j and I got t=3 and v=2. I also got the angle to be 33.69°. I don't know what to do after that of if anything i got is right.

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Is that the way the problem statement is stated your textbook?

If particle $A$ is only moving along the $x-axis$ and particle $B$ is only moving along the $y-axis$, their relative position is always going to be the hypotenuse of a right triangle with a side being a multiple of $3$ and another being a multiple of $v$. Can you figure out what the expression for the position of $B$ relative to $A$ for all $t$ is given their relative position at $t=0$?

Yes, that is the way the problem is stated.

So the relative position is vj-3i ?

haruspex
Homework Helper
Gold Member
So the relative position is vj-3i ?
No, that's the relative velocity. You were right to integrate that (I assume v is constant).
But I'm not sure what you meant by this:
I integrated A and B and let it equal to -9i+6j and I got t=3
When you integrate you get an unknown constant, and you have to find the value of that from initial conditions. What general formula did you get for relative position?

When I integrate A= 3i , I will get 3ti + constant and B= vj so that it will be vtj + constant. I added the t because, the answer has to be with respect to time. And then I let it equal to -9i+6j= -3ti +vtj

So you mean that relative position is basically + vtj -3ti because its B relative to A.

And from that I can say that v has to be 2 for them to collide, but what about part 3?

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haruspex