O-LevelAdditional MathematicsVectors in two dimensions4037/12 · May/June 2024 · Q910 Marks

In this question, all distances are in metres and time, t, is in seconds.

(a)A particle P moves with a speed of 14.5 parallel to the vector (-20, 21). Find the velocity vector of P.[2]
(b)Initially, P has position vector (3, 5). Write down the position vector of P at time t.[2]
(c)A second particle Q has position vector (-1, 3) + (-5, 7.5)t at time t. Find, in terms of t, the distance between P and Q at time t. Simplify your answer.[4]
(d)Hence show that P and Q never collide.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

[B2] (-10, 10.5)
B1 for |-20, 21| = 29 soi or one correct element
[2] (3, 5) + (-10, 10.5)t oe FT on their velocity vector from part (a) if not correct.
[M1] PQ = ( (5t-4), (-3t-2) ) oe Dep M1 must have M1 in part (b) for P_Q = ( (-10t+3) - (-5t-1), (10.5t+5) - (7.5t+3) ) - their answer to part (b) or their answer to part b ( (-1, 3) + (-5, 7.5)t ) - ( (3, 5) + (-10, 10.5)t )
[A1] PQ = √( (5t-4)² + (-3t-2)² )
[M1] PQ = √( 34t² - 28t + 20 ) Dep on previous M mark for using their PQ
[A1] √( 34t² - 28t + 20 )
[M1] For collision 34t² – 28t +20=0 soi Need to be using their quadratic PQ² from part (c) or forming a quadratic from their PQ from part (c)
[A1] Discriminant = -1936 So no real roots so no collision oe Discriminant must be correct, but allow unsimplified

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