A particle is projected with velocity vector (10i + 15j)ms^-1 where i and j are unit vectors in the horizontal and upward vertical directions respectively. Find its velocity vector 2 seconds later and its distance from the point of projection at this time. Well first of all, could anyone explain the difference between the velocity vector after 2 seconds, and the position vector after 2 seconds? working: I got 20i + 10.4j for the position vector after 2 seconds, using pythagoras I was able to find the distance from the point of projection (22.5m), for the velocity vector I didn't really know what to do, so I looked at the answer and deduced how they got it. So using the initial velocity vector (10i + 15j) as a point O, and the new position (20i + 10.4j) I manage to get the velocity vector to be (20-10)i + (15-10.4)j so 10i + 4.6j, however the answer is 10i - 4.6j Am I wrong, or is the book? If further proof of working is needed I'll scan the paper I used. Thanks.