A cyclist, when travelling due west at 15 km h\(^{-1}\), finds that the wind appears to be blowing from a bearing of 150°. When the cyclist is travelling due west at 10 km h\(^{-1}\), the wind appears to be blowing from a bearing of 135°. Find the velocity of the wind. [10]
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Question 12:
Answer Marks
12 Method α: D 10 B 5 A
θ 135° 120°
∠ABC = 45°
∠BAC = 120° w v u
∠BCA = 15°
C
v 5 5sin120°
= ⇒v= =16.73
sin120° sin15° sin15°
w 2 = 10 2 + 16.73 2 – 2 × 10 × 16.73 × cos 135 o
⇒ w = 24.8(3)
sinθ sin135°
=
16.73 24.83
⇒ θ = 28.45°.
⇒ Wind blows from bearing 118°.
−sin30° −15 −sin45° −10
Method β: u + =w +
cos30° 0 cos45° 0
u w
+ 15 = + 10
2 2
3u w
=
2 2
5 ( )
u = 5 + 5 3 = 13.66 or v = 2 3+ 3 =9.66
2
5( )
− 7+ 3 −21.83
2
=
Either side =
5( )
3+ 3 11.83
2
wsinθ
Method γ: 15+ =wcosθ
3
10 + w sin θ = w cos θ
9+2 3
tanθ = ⇒ θ = 28.45°.
23
Wind blows from bearing 118°.
–1
Answer Marks
v = 24.8(3). Speed of wind is 24.8 km h . Diagram, with all three
u 5
Or: =
sin135o sin15o
5sin135o
⇒u = =13.66or5+5
sin15o
(N.B. Accept this vector answer
from any method)
[Note: If a and b for wsinθ and
wcosθ in the above equations,
each equation gets M1A1,
followed by M1A1A1 for v and
Answer Marks
M1A1A1 for θ.] B1
M1A1
A1
M1A1
A1
M1
A1
A1 [10]
M1A1
M1A1
M1
M1
A1
M1A1
A1 [10]
M1A1A1
M1A1A1
M1A1
M1A1[10]
Answer Marks
Guidance
Page 8 Mark Scheme
Syllabus
Pre-U – May/June 2014 9795
02
Method δ:
w 15 15sin120°
= ⇒w=
( )
sin120° sin 60°−θ sin60°cosθ −cos60°sinθ
15 3
⇒w=
3cosθ −sinθ
w 10 10sin135°
= ⇒w=
( )
sin135° sin 45°−θ sin45°cosθ −cos45°sinθ
10
⇒w
cosθ −sinθ
Equate: 15 3 cosθ – 15 3 sinθ = 10 3 cosθ –10 sinθ
5 3 9+2 3
⇒tanθ = = [= 0.5419]
15 3−10 23
⇒ θ = 28.45 ⇒ Wind blows from bearing 118°
and w = 24.83
Method ε : Components:
w x = u x + 15 = v x + 10
w y = u y = v y
w w
y =tan ( 60° ) = 3, y =tan ( 45° ) =1
w −15 w −10
x x
5
w x – 10 = 3(w x – 15) ⇒ w x = 2 (7+ 3),
5
w = (3+ 3)
y 2
⇒ speed = 24.8(3) ms –1
Angle is tan –1 (0.5419) = 28.45°
Answer Marks
⇒ wind blows from bearing 118°. Two equations for θ and w
Compound angle formulae used
Correct values of sin and cos
Solve
Correct θ; correct bearing
Correct w
Any two
M1A1
M1A1
M1
A1A1
A1 [10]
M1
A1
M1
A1
A1
M1A1
A1
M1
A1 [10]
5( )
− 7+ 3 −21.83
2
Summary of answers: w = = ,w=2 16+5 3 =24.83
5( )
3+ 3 11.83
2
3+ 3 18+4 3
θ =tan−1 or tan−1 =28.45°⇒wind blows from bearing 118(.45)°
7+ 3 46
5( ) 5( )
− 1+ 3 −6.83 − 1+ 3 −11.83
2 2
u = = ,u=13.66 v = = ,v=16.73
5( 3+ 3 ) 11.83 5( 3+ 3 ) 11.83
2 2
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Question 12:
12 | Method α: D 10 B 5 A
θ 135° 120°
∠ABC = 45°
∠BAC = 120° w v u
∠BCA = 15°
C
v 5 5sin120°
= ⇒v= =16.73
sin120° sin15° sin15°
w 2 = 10 2 + 16.73 2 – 2 × 10 × 16.73 × cos 135 o
⇒ w = 24.8(3)
sinθ sin135°
=
16.73 24.83
⇒ θ = 28.45°.
⇒ Wind blows from bearing 118°.
−sin30° −15 −sin45° −10
Method β: u + =w +
cos30° 0 cos45° 0
u w
+ 15 = + 10
2 2
3u w
=
2 2
5 ( )
u = 5 + 5 3 = 13.66 or v = 2 3+ 3 =9.66
2
5( )
− 7+ 3 −21.83
2
=
Either side =
5( )
3+ 3 11.83
2
wsinθ
Method γ: 15+ =wcosθ
3
10 + w sin θ = w cos θ
9+2 3
tanθ = ⇒ θ = 28.45°.
23
Wind blows from bearing 118°.
–1
v = 24.8(3). Speed of wind is 24.8 km h . | Diagram, with all three
u 5
Or: =
sin135o sin15o
5sin135o
⇒u = =13.66or5+5
sin15o
(N.B. Accept this vector answer
from any method)
[Note: If a and b for wsinθ and
wcosθ in the above equations,
each equation gets M1A1,
followed by M1A1A1 for v and
M1A1A1 for θ.] | B1
M1A1
A1
M1A1
A1
M1
A1
A1 [10]
M1A1
M1A1
M1
M1
A1
M1A1
A1 [10]
M1A1A1
M1A1A1
M1A1
M1A1[10]
Page 8 | Mark Scheme | Syllabus | Paper
Pre-U – May/June 2014 | 9795 | 02
Method δ:
w 15 15sin120°
= ⇒w=
( )
sin120° sin 60°−θ sin60°cosθ −cos60°sinθ
15 3
⇒w=
3cosθ −sinθ
w 10 10sin135°
= ⇒w=
( )
sin135° sin 45°−θ sin45°cosθ −cos45°sinθ
10
⇒w
cosθ −sinθ
Equate: 15 3 cosθ – 15 3 sinθ = 10 3 cosθ –10 sinθ
5 3 9+2 3
⇒tanθ = = [= 0.5419]
15 3−10 23
⇒ θ = 28.45 ⇒ Wind blows from bearing 118°
and w = 24.83
Method ε : Components:
w x = u x + 15 = v x + 10
w y = u y = v y
w w
y =tan ( 60° ) = 3, y =tan ( 45° ) =1
w −15 w −10
x x
5
w x – 10 = 3(w x – 15) ⇒ w x = 2 (7+ 3),
5
w = (3+ 3)
y 2
⇒ speed = 24.8(3) ms –1
Angle is tan –1 (0.5419) = 28.45°
⇒ wind blows from bearing 118°. | Two equations for θ and w
Compound angle formulae used
Correct values of sin and cos
Solve
Correct θ; correct bearing
Correct w
Any two
All four | M1A1
M1A1
M1A1
M1
A1A1
A1 [10]
M1
A1
M1
A1
A1
M1A1
A1
M1
A1 [10]
5( )
− 7+ 3 −21.83
2
Summary of answers: w = = ,w=2 16+5 3 =24.83
5( )
3+ 3 11.83
2
3+ 3 18+4 3
θ =tan−1 or tan−1 =28.45°⇒wind blows from bearing 118(.45)°
7+ 3 46
5( ) 5( )
− 1+ 3 −6.83 − 1+ 3 −11.83
2 2
u = = ,u=13.66 v = = ,v=16.73
5( 3+ 3 ) 11.83 5( 3+ 3 ) 11.83
2 2
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A cyclist, when travelling due west at 15 km h$^{-1}$, finds that the wind appears to be blowing from a bearing of 150°. When the cyclist is travelling due west at 10 km h$^{-1}$, the wind appears to be blowing from a bearing of 135°. Find the velocity of the wind. [10]
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2014 Q12 [10]}}