Short-Term Conflict Resolution for Unmanned Aircraft Traffic Management
1. A has a forward speed 2of 18 km/h and is accelerating at 3 m/s...
Transcript of 1. A has a forward speed 2of 18 km/h and is accelerating at 3 m/s...
1. The car A has a forward speed of 18 km/h and is accelerating at 3 m/s2.
Determine the velocity and acceleration of the car relative to observer B, who
rides in a nonrotating chair on the Ferris wheel. The angular rate
Ω = 3 rev/min of the Ferris wheel is constant.
velocity and acceleration of the car relative to observer B: BABABABA aaavvv −=−= //
iv
smsmhkmv
A
A
5
/5/6.3
18/18
=
===
iasma
A
A 3
/3 2
=
=
( )
( ) jijivv
smRvsradsradrev
BB
B
2245sin45cos
/827.2910
/10
/6023min/3
827.2
−=−=
==Ω=⇒===Ωπππ
( ) ( ) jijiaasmRvRa BB
BB
628.0628.045sin45cos/888.0910
888.0
222
2 −−=−−=⇒=
==Ω=
π
Velocity of A with respect to B
( ) jivjiivvv BABABA
23225 // +=⇒−−=−=
Acceleration of A with respect to B
( ) jiajiiaaa BABABA
628.0628.3628.0628.03 // +=⇒−−−=−=
AaAv
BvBa
2. The aircraft A with radar detection equipment is flying horizontally at an altitude
of 12 km and is increasing its speed at the rate of 1.2 m/s each second. Its radar
locks onto an aircraft B flying in the same direction and in the same vertical plane
at an altitude of 18 km. If A has a speed of 1000 km/h at the instant when θ=30o,
determine the values of and at this same instant if B has a constant speed of
1500 km/h.
r θ
3. Airplane A is flying horizontally with a constant speed of 200 km/h and is
towing the glider B, which is gaining altitude. If the tow cable has a length r =60 m
and θ is increasing at the constant rate of 5 degrees per second, determine the
magnitudes of the velocity and acceleration of the glider for the instant when
θ=15o.
vA=200 km/h (cst), r =60 m,
when θ=15o.
)(/180
5deg/5 cstsrads πθ ==
ABABABBAAB vvvrrrrrr //
+=+=+=?? == BB av
O
Br
Ar
rr AB
=/
r+
θ+
iv
smsmhkmv
A
A
55.55
/55.55/6.3
200/200
=
===
θθererrv rAB
+==/
( ) 060 =⇒= rcstmr
θπθ evsmr AB
23.5/23.5180
560 / =⇒=
=
(in polar coordinates)
jiv AB
15cos23.515sin23.5/ += (in cartesian coordinates)
Velocity of the glider:
jivjiivvv
B
ABAB
05.5904.56
15cos23.515sin23.555.55/
+=
++=+=
θ=15o
θ=15
o
θ=15o
vA=200 km/h (cst), aA=0 )(/180
5deg/5 cstsrads πθ ==
ABAB aaa /0
+=
?=Ba
O
Br
Ar
rr AB
=/
r+
θ+( ) ( ) θθθθ errerrraa rABB
22/ ++−===
rrrB eeera 456.0180
5602
2 −=
−=−=
πθ
(in polar coordinates: From B to A)
Acceleration of the glider:
θ=15o
θ=15
o
θ=15o )( cst=θ)( cstr = )( cstr =
4. Particles A and B both have a speed of
8 m/s along the directions indicated
arrows. A moves in a curvilinear path
defined by y2=x3 and B moves along a
linear path defined by y=−x. If the
velocity of B is decreasing at a rate of 6
m/s each second and the velocity of A is
increasing at a rate of 5 m/s each second,
determine the velocity and acceleration of
A with respect to B for the instant
represented.
vA=vB=8 m/s. Velocity of B is decreasing at a rate of 6 m/s each second and velocity of A is increasing at a rate of 5 m/s each
second. BABABABA aaavvv
−=−= //
Velocity of A
o
x
xdxdy
xyxy
31.5623
23tan 2/1
1
2/332
=⇒===
=⇒=
=
θθ
Av
Bv
jivjiv
A
A
656.6437.431.56sin831.56cos8
+=
+=
Avθ
jivvv BABA
312.12219.1/ +−=−=
θ
Velocity of B
jivjiv
B
B
656.5656.545sin845cos8
−=
−=
Velocity of A wrt B
450
1 m
Velocity of B is decreasing at a rate of 6 m/s each second and velocity of A is increasing at a rate of 5 m/s each second.
Acceleration of A (Curvilinear Motion)
m
dxyd
dxdy
812.7
432311
2/32
2
2
2/32
=
+
=
+
=ρ
Av
Bv ( )( )
( )( )
jia
jijia
A
aa
A
tAnA
704.8043.4
31.56sin531.56cos569.33sin192.869.33cos192.8
+−=
+++−=
o31.56=θ
jiaaa BABA
461.42.0/ +=−=
θ
Acceleration of B (Rectilinear Motion)
Acceleration of A wrt B
450
1 m
( ) ( )nAtAA aaa +=
+t +n
o69.33
+t
( )nAa ( )tAa ( ) smvva AA
nA /82
==ρ
43,
23
23
2
22/1
1
==== dx
ydxdxdy
x
( ) ( ) 222
/5/192.8812.78 smasma tAnA ===
Ba
jiajia BB
243.4243.445sin645cos6 +−=⇒+−=
60 m
30o
30 m A
B
θ r
60 m 30o
30 m
A
B
θ r
30o
+r
+θ
+t
+n
Bv
BaAv
Aaα
β
β
60 m
30o
30 m A
B
θ r
BABA vvv −=/
6. A batter hits the baseball A with an initial velocity of vo=30 m/s directly toward fielder B at an angle of 30° to the horizontal; the initial position of the ball is 0.9 m above ground level. Fielder B requires sec to judge where the ball should be caught and begins moving to that position with constant speed. Because of great experience, fielder B chooses his running speed so that he arrives at the “catch position” simultaneously with the baseball. The catch position is the field location at which the ball altitude is 2.1 m. Determine the velocity of the ball relative to the fielder at the instant the catch is made.
41
( ) ( ) jivijivvv BABABA
19.1446.2152.419.1498.25 // −=⇒−−=−=
x
y
( ) ( ) ( )ss
tttgttvyy yoo 08.098.2
81.92130sin309.01.2
21
2,122 =⇒−+=⇒−+=
C «Catch position»
2.1 m
Total flight time
Range: ( ) ( )( ) mxtvxx xoo 3.7798.230cos30 ==⇒+=
m3.12653.77 =− (Fielder B must run 12.3 m in 2.73 seconds.)
s73.24198.2 =− (the duration that fielder B must run)
Velocity of fielder B: ivcstsmv BB
52.4)(/52.473.23.12
=⇒==
Velocity components of the ball at C:
( ) smvv xox /98.2530cos30 ===in x-direction:
in y-direction: ( ) ( ) smgtvv yoy /19.1498.281.930sin30 −=−=−=
jivA
19.1498.25 −=
Velocity of the ball relative fielder
7. Small wheels attached to both ends of rod AB roll along smooth surfaces.
At the instant shown, the wheel A has a velocity of 1.5 m/s towards right and
its speed is increasing at a rate of 0.5 m/s2. Determine the absolute velocity of
wheel B ( ), its relative velocity with respect to A ( ), its absolute
acceleration ( ) and its relative acceleration with respect to A ( ).
800 mm
500 mm A
B 60°
Bv A/Bv
Ba A/Ba
Av
800 mm
500 mm
A
B
60°
Av
Bv
x
y In Cartesian Coordinates
In Polar Coordinates
800 mm
500 mm
A
B
60°
Av
Bv+r
+θ
θ
β
β 30° θ
120°
θ
A/BBB
A/BAB
BBB
BBB
A
vi.jv.iv.vvv
jv.iv.vjsinvicosvv
i.v
+=−
+=−=
−=
=
51866050
8660506060
51
800 mm
500 mm
A
B
60°
Av
Bv+r
+θ
θ
β
β 30° θ
120°
θ
r=0.8 m
25277532120
800500 ..sinsin
=== θββ
( ) ( )θθθθ
θθ
θθ
θ
θ
ββ
θθ
e.e.ev.ev.evev)rrconstantmmr(rv
evevvvv
ev.ev.esinvecosvve.e.v
esinvecosvv
rBrBrr
r
rrABA/B
BrBBrBB
rA
ArAA
687033315410841008000
5410841068703331
0
−−+=+==−===
+=−=
+=+=−=
−=
Velocity:
j.i.jcos.isin.ve.v
s/rad..
.s/m.vs/m.v
.v.ve.v.e
A/B
A/B
B
B
Br
3731707054415441
5441
93180
544154415851
68705410333184100
−−=−−=
=
====
+=→−=→
θθ
θ
θ
θ
θθ
800 mm
500 mm
A
B
60°
Av
Bv+r +θ
θ
β
β 30° θ
120°
θ
800 mm
500 mm
A
B
60°
Aa
Ba+r
+θ
θ
β
β 30° θ
120°
θ
( )( ) ( )θθθθ
θθ
θθ
θ
θ
θ
ββ
θθ
e.e.ea.ea.eaeas/m...rra
eaeaaaa
ea.ea.esinaecosaae.e.a
s/m.aesinaecosaa
rBrBrr
r
rrABA/B
BrBBrBB
rA
AArAA
230444054108410982931800
541084102304440
50
222
2
−−+=+−=−=−=
+=−=
+=+=−=
=−=
Acceleration:
( ) ( )θθθθ e.e.ea.ea.eaea rBrBrr 230444054108410 −−+=+
θ
θ
θθ
e.e.a
s/m.as/m.a
.a.ae
.a.ae
rA/B
B
B
B.rr
41982
4110153
2305410
44408410
22
982
−−=
−=−=
−=→
−=→−