Mathematics · May / June 2023 · Paper 51 · Question 8
YMA01
As printed in the paper
The question as text
8. 50 m D C 50 m O A B Figure 5 A square floor space ABCD, with centre O, is modelled as a flat horizontal surface measuring 50 m by 50 m, as shown in Figure 5. → and AD → respectively. The horizontal unit vectors i and j are in the direction of AB All position vectors are given relative to O. A small robot R is programmed to travel across the floor at a constant velocity. At time t = 0 , R is at the point with position vector (–2i + j) m • At time t = 11 s , R is at the point with position vector (9i + 23j) m • At time t seconds, the position vector of R is r metres •
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Show the answer and mark scheme
8(a)[3]
v
=
11
Expression for r with correct structure
r
i
j
i
j or
r
i
j
= −
( 2
+
)
+
t
(
+
2 )
=
(
t
−
2)
+
(2
t
+
1)
s
=
(25
i
+
25 )
j
+ −−
t
(
i
j
)
8(b)[1]
Or
s
=
(25
−
t
)
i
+
(25
−
t
)
j
8(c)[2]
Either −
s
r
Or −
s
r
with their r and s substituted
[
]
SR
=
(2
t
−
27)
i
+
(3
t
−
24)
j
m *
8(d)[3]
Distance
( )
d
=
(2
t
−
27)
2
+
(3
t
−
24)
2
2
2
2
(
d
)
=
(2
t
−
27)
+
(3
t
−
24)
(
d
2
)
=
13
t
2
−
252
t
+
1305
126
t =
=
9.7
(s) or better
13
Notes for Question 8
(a)
M1
Use of displacement/time to find velocity. Allow the difference either way round.
Expression for r with correct structure using their v and the correct initial position
M1
vector.
A1
Correct expression in terms of t, i and j
(b)
Any correct expression for s in terms of t, i and j
B1
(c)
M1
(Their r – their s) or vice versa, unsimplified
A1*
Correct answer correctly obtained. Allow missing square brackets and m, but rest
must be identical to given answer.
(d)
M1
Use of Pythagoras to find an expression for distance (or distance squared)
A1
Correct 3 term quadratic expression
N.B. If no 3 term quadratic expression is seen but a correct derivative is, award this
mark.
A1
9.7 or better.
N.B. If a fraction is given as the answer, it must be the ratio of two positive integers
or a mixed fraction.
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