Mathematics · May / June 2022 · Paper 51 · Question 8
YMA01
As printed in the paper
The question as text
8. [In this question, i and j are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin O.] Two boats, P and Q, are moving with constant velocities. The velocity of P is 15i m s–1 and the velocity of Q is (20i – 20j) m s–1
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[2]
isosceles triangle.
M0 if not using the velocity of Q
135o
i
8(b)[5]
p
=
t
(15 )
=
200
+
(20
−
20 )
oe
q
j
t
i
j
=
−
=
200
+
(
20
−
20
)
−
(15 )
PQ
q p
j
t
i
j
t
i
(M0 if they put p = q)
Need to see at least this line of working.
5
(200
20 )
PQ
=
t
i
+
−
t
j
(m) *
PQ =
50
i
at t = 10
8(c)[2]
(Allow M1 if they find p = 150i and q = 200i at t = 10)
N.B. This mark could be implied by a correct diagram
270O
8(d)[5]
5t = 200 − 20t
t = 8
(5 8)
(200 20 8)
PQ =
×
i
+
−
×
j
PQ =
40
2
+
40
2
=
40 2
(m) , 57 or better
(5 )
2
+
(200 20 )
−
2
=
200
2
8(e)[3]
t
t
425
2
−
8000
=
0
t
t
320
t = 0 or
17 =18.82… Accept 19 or better Apply isw
Notes for question 8
8(a)
M1 e.g. use of cos or sin or the cosine rule
A1 cao
M1 Correct structure for either
8(b)
A1 oe
A1 oe
M1 Allow p − q
A1* Correct given answer correctly obtained, allow omission of m.
8(c)
M1 Clear attempt to find PQ
at t = 10 or p and q at t = 10
A1 cao
M1 Equating components to give an equation in t only, with no vectors
8(d)
A1 cao
M1 Substituting their t value into their PQ
M1 Finding the magnitude, with square root
A1 cao
M1 Use of Pythagoras – allow with square root – to obtain an equation
8(e)
in t.
A1 Correct unsimplified 2 term quadratic in t.
A1 For both answers
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