# Difficult vector question

## Homework Statement

Particle A moves along the line y=30m with a constant velocity vector of magnitude 3/ms and it is parallel to the x axis. Right when particle A passes the y-axis Particle B leaves the origin with 0 initial speed and a constant acceleration magnitude of .4m/s/s. What angle theta (which is angle between the acceleration vector and y axis) would result in a collision, also particle B travels in a linear path as well.

## Homework Equations

x-xo=(Vocos(theta))t+.5a(cos(theta))t^2
(I believe that is it)

## The Attempt at a Solution

My thoughts on this problem are to make (Vocos(theta))t+.5a(cos(theta))t^2=(Vocos(theta))t+.5a(cos(theta))t^2, or their x displacements the same, but then again I think I should take account for the y for Particle B, but I'm not exactly sure how to approach this problem though...

You are on the right track
A diagram will help, in this case a x-y displacement graph
It shows the movement of the particle A parallel to the x axis at a distance 30m
It shows the linear path of B at an angle θ such that they meet at a point
Let this happen after a time t seconds
Then the distance A moves is 3t and the distance B moves is ut + ½at² = ½at²
Use the triangle to find t and θ

Hint: Pythagoras!