This the Advance Math Project for G12 ES, Term 1.

3. Group Members

4. THE HISTORY OF AIRBUS
A380
• The 555 seat, double deck Airbus A380 is the world's
largest airliner, easily eclipsing Boeing's 747. The A380 base
model is the 555 seat A380-800 (launch customer
Emirates). Potential future models include the 590 ton
MTOW 10,410km (5620nm) A380-800F freighter, able to
carry a 150 tonne payload, and the stretched, 656 seat,
A380-900.
• Airbus first began studies on a very large 500 seat airliner in
the early 1990s. The European manufacturer saw
developing a competitor and successor to the Boeing 747
as a strategic play to end Boeing's dominance of the very
large airliner market and round out Airbus' product line-up.
• Airbus began engineering development work on such an
aircraft, then designated the A3XX, in June 1994.

5. Airbus A380 characteristics.
http://www.slideshare.net/rubal_9/airbus-a380-
1856096

6. How long does it take to take off, and
the velocity of it
http://www.youtube.com/watch?v=p4DTormtE
G0

7. What is the take o f f distance , i . e. t h e runway length that will
be crossed before it takes off ?
• Generally, 8000 feet is long enough for
anything that flies nowadays. the a380 was
designed, as much as possible, to be
compatible with 747 infrastructure.

9. Speed of an airplane by Radar
• X2+Y2=52
X2+62=52
X2+62=(10)2
X2=102-62=64
X=√64
=8
X2+Y2=52
2X×(dx/dt)+2y×(dy/dt)=2s×(ds/dt)
*2y×(dy/dt)=0 because the height don’t change
2(8)×(dx/dt)=2(10)×(400)
(dx/dt)=8000/16
=500

11. Waiting at the airport
• ω(x)=X2/2(1-X)
ω|(x)=(2-2x)(2x)-X2(2)×(-1)/4(1-x)2
ω|(x)=-4x2+4x+2x2/4(1-x)2
ω|(x)=-2x2+4x/4(1-x)2
=8x(2-x)/4(1-x)2
=x(2-x)/2(1-x)2
ω|(0.1)=(0.1)×(2-0.1)/2(1-0.1)2
=0.19/1.62
=0.1173
ω|(0.7)=(0.7)×(2-0.7)/2(1-0.7)2
=0.91/0.18
=5.0556
•

13. Taking off Distance and Time
• D(t)=(10/9)t2
D|(t)=20/9 ×t=72
t=72/(20/9)=32.4s
D(32.4)=(10/9)(32.4)2=1166.4m

15. A.G.P
Part a
• When x=-4 , y=1
• F(x)=9x3+bx2+cx+d
• Also, when x=0, y=0 when hitting the origin, 50 , d=0
• 50, f(X)=9x3+bx2+cx+0
• 1=-64a+16b-4c
• -1=64a-16b+4c
• f ’(x)=3ax2+26x+c ; -4,-1
• f ’(-4)=0 , f ’(0)=0
• When x=0 then 3(0)+2(0)+c=0
c=0

16. A.G.P
Part a
• When x=-4, 48a-8b=0
48a=8b
6a=b
• From before , 64a-16b+4(0)=-1
64a-16(6a)=-1
64a-96a=-1
-32a=-1
a=1/32
• b=6a=6x1/32=3/16
F(x)=1/32x3+3/16x2

17. A.G.P
Part b
• f ’(x)=3/32x2+3/8x
f ’’(x)=3/16x+3/8=0
3/16x=-3/8
X=-2
• When x=-2
f(-2)=3/32(-2)2+3/8(-2)
=(3/8)-(3/4)
=-3/8
• When x=-2
f(-2)=1/32(-2)3+3/16(-2)2
=-1/4+3/4
=1/2
So the max rate would be at (-2,1/2)

19. • F is frequency perceived by the observer.
• fs is the frequency of the source.
• Vw is the velocity of the waves.
• VL is the velocity of the listener.
• Vs the velocity of the source.

20. • A) Find an equation of the F in term of Vs.
• I Will assume that this is a A380 Airbus.
• V = speed of waves (sound) = 340 m/s
• VL = 0, since he isn’t moving.
• Where L is the length of the airbus and g is the gravity of earth
• Therefore,
• Which is :

21. • b)
• c)