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

−−=

−=−=

−=→

−=→−