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# M1 vectors, last part of question watch

1. Tomorrow is M1 day and i've been doing past papers and got stuck at part d) of this question:

M1 January 2006 paper:
6.
[In this question the horizontal unit vectors i and j are due east and due north
respectively.]
A model boat A moves on a lake with constant velocity (–i + 6j) m s–1.
At time t = 0, A is at the point with position vector (2i – 10j) m. Find:

(a) the speed of A,
(2)
(b) the direction in which A is moving, giving your answer as a bearing.
(3)
At time t = 0, a second boat B is at the point with position vector
(–26i + 4j) m.

Given that the velocity of B is (3i + 4j) ms-1

(c) show that A and B will collide at a point P and find the position vector of P.

Given instead that B has speed 8 m s–1 and moves in the direction of the vector (3i + 4j),

(d) find the distance of B from P when t = 7 s.

I get everything until part d), I don't understand how I would turn the speed given into a velocity to be able to work out the position vector, and then the distance.

many thanks
2. im pretty sure u need to work out the PV'S of both when t= 7 then minus them
3. well speed is the magnitude of velocity, so you can find the magnitude of your direction vector by doing the square root of (3squared + 4squared). this'll give 5, and divide 8/5 to get the number of times you need to multiply the direction vector to get the right speed.

sooo the velocity will be 1.6(3i + 4j), and i think you can work it out from there.
4. (Original post by BrightGirl)
well speed is the magnitude of velocity, so you can find the magnitude of your direction vector by doing the square root of (3squared + 4squared). this'll give 5, and divide 8/5 to get the number of times you need to multiply the direction vector to get the right speed.

sooo the velocity will be 1.6(3i + 4j), and i think you can work it out from there.
wooo i understand!
thanks
5. (Original post by bob'syouruncle)
wooo i understand!
thanks
most welcome
6. (Original post by BrightGirl)
well speed is the magnitude of velocity, so you can find the magnitude of your direction vector by doing the square root of (3squared + 4squared). this'll give 5, and divide 8/5 to get the number of times you need to multiply the direction vector to get the right speed.

sooo the velocity will be 1.6(3i + 4j), and i think you can work it out from there.
i know its 2years late but any chance of anyone explaining the bold bit to me in a bit more detail?
7. is this OCR (not MEI) or is this MEI??
8. (Original post by BrightGirl)
well speed is the magnitude of velocity, so you can find the magnitude of your direction vector by doing the square root of (3squared + 4squared). this'll give 5, and divide 8/5 to get the number of times you need to multiply the direction vector to get the right speed.

sooo the velocity will be 1.6(3i + 4j), and i think you can work it out from there.
Sorry I stiil don’t get the part that you need to divide it by 8/5
9. (Original post by Maggie07)
Sorry I stiil don’t get the part that you need to divide it by 8/5

Make a new thread with your question if you need help with it.
10. (Original post by RDKGames)

Make a new thread with your question if you need help with it.
thank you so much😊

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