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Question
A particle P is initially at the point (2,6) in relation to an origin, O, and is moving with velocity (3i +j) mps.  It has constant acceleration (16i + 24j) mpsps.  Show that after 2 seconds it is moving directly away from O, and find its speed at that time.

In the above question I have found that its velocity is 35i + 49j and its speed is 60.2.  However, I don't know what it means by "show that after 2 seconds it is moving directly away from O".

The book tells me in the answers that position r = (40i + 56j)m but I'm no wiser and I can't get that value at all and even when I have that information I don't know what to do with it.

Any help would be much appreciated.

Thanks

Answer
Hi Maggie,
When a particle moves on a path we know that a tangent line at any point is in the direction of the velocity. Moving directly away from the origin simply means that this tangent line passes through the origin. To visualize it, imagine that another particle leaves the origin and is moving in a certain straight line such that it would pass through the said point on the path of the previous particle. At the instant when each particle is at that point they would both be momentarily headed in the same direction.
Vectorially, it just means that the unit vector in the direction of the velocity of particle P at that point would be the same as the unit vector of the position vector of the point.
Now, you can get the position vector by using any of the simple equations of motion with constant acceleration. We could use
Δs = ut + ½at²
So, ΔR = (3i + j).2 + ½(16i + 24j).2²
ΔR = (6i + 2j) + (32i + 48j)
ΔR = 38i + 50j
and
R = (2i + 6j) + (38i + 50j)
R = 40i + 56j
All that is left to do now is compare the unit vectors, which are
(40i + 56j)/√(40² + 56²) and (35i + 49j)/√(35² + 49²)
which i'm sure you'll find to be the same.
Always remember to fully understand a question and not just the mathematics involved, glad you've tried to do that here.

Regards

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