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:

aB
18
v A = 18 km / h =
m/s = 5 m/s
3.6


v A = 5i

vB

aA

vA
π
2π
Ω = 3 rev / min = 3
rad / s =
rad / s
60
10





vB = vB (cos 45i − sin 45 j ) = 2i − 2 j

2.827
vB2  π 
2
aB = Ω R =
=   (9 ) = 0.888 m / s 2
R  10 
2
Velocity of A with respect to B




 
v A / B = v A − vB = 5i − (2i − 2 j ) ⇒

  
 
v A / B = v A − vB a A / B = a A − aB
⇒
aA = 3 m / s2


a A = 3i
⇒
vB = ΩR =
π
10
(9) = 2.827 m / s





aB = aB (− cos 45i − sin 45 j ) = −0.628i − 0.628 j
0.888



v A / B = 3i + 2 j
Acceleration of A with respect to B




 
a A / B = a A − aB = 3i − (− 0.628i − 0.628 j )
⇒



a A / B = 3.628i + 0.628 j
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 r and θ at this same instant if B has a constant speed of
1500 km/h.
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.
θ=15o
π
rad / s (cst )
vA=200 km/h (cst), r =60 m, θ = 5 deg/ s = 5
180



  


o
vB = ? aB = ? when θ=15 .
rB = rA + rA / B {rB = rA + r }
+r
+θ
θ=15o





vB / A = r = rer + rθeθ
r = 60 m (cst ) ⇒
r = 0
θ=15o

rA
O
200
v A = 200 km / h =
m / s = 55.55 m / s
3.6


v A = 55.55i


rB / A = r

rB

 
vB = v A + vB / A


 π 
5
.
23
/
5
.
23
rθ = 60 5
m
s
v
e
=
⇒
=

θ
B/ A
 180 
(in polar coordinates)



vB / A = 5.23 sin 15i + 5.23 cos15 j (in cartesian coordinates)




 
Velocity of the glider: vB = v A + vB / A = 55.55i + 5.23 sin 15i + 5.23 cos 15 j



vB = 56.904i + 5.05 j
θ = 5 deg/ s = 5
θ=15o
vA=200 km/h (cst), aA=0
+r
+θ
π
180

aB = ?
rad / s (cst )

 
aB = a A + aB / A

0





aB = aB / A = r = r − rθ 2 er + rθ + 2rθ eθ
(
) (
( r = cst )
θ=15o
)
(θ = cst ) ( r = cst )


rB / A = r

rB
θ=15o

rA
Acceleration of the glider:



 π  
aB = −rθ 2 er = −60 5
e
0
.
456
e
=
−
 r
r
180


2
O
(in polar coordinates: From B to A)
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

  
 
v A / B = v A − vB a A / B = a A − aB
second.

vA

vA
Velocity of A
y 2 = x3 ⇒
θ
tan θ =
dy
dx
θ
y = x3/ 2
=
x =1
3 1/ 2 3
x =
2
2
⇒



v A = 8 cos 56.31i + 8 sin 56.31 j



v A = 4.437i + 6.656 j
1m
450
Velocity of B

vB
Velocity of A wrt B
θ = 56.31o



vB = 8 cos 45i − 8 sin 45 j



vB = 5.656i − 5.656 j



 
v A / B = v A − vB = −1.219i + 12.312 j
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.

vA
+t
θ
+n
(a A )n
33.69
450

aB
dy
dx
1m

vB



a A = (a A )t + (a A )n
Acceleration of A (Curvilinear Motion)
+t
  dy  2 
1 +   
  dx  
ρ=
d2y
dx 2
82
=
= 8.192 m / s 2
7.812
3/ 2
  3 2 
1 +   
  2  
=
3
4
3/ 2
= 7.812 m





a A = (− 8.192 cos 33.69i + 8.192 sin 33.69 j ) + (5 cos 56.31i + 5 sin 56.31 j )


 
( a A )t



a A = −4.043i + 8.704 j
(Rectilinear Motion)
vA = 8 m / s
ρ
(a A )t = 5 m / s 2
( a A )n
Acceleration of B
v A2
θ = 56.31o
o
3 1/ 2 3 d 2 y 3
= x = ,
=
2
2
2
dx
4
x =1
(a A )n
(a A )n =
(a A )t



aB = −6 cos 45i + 6 sin 45 j
Acceleration of A wrt B
⇒



aB = −4.243i + 4.243 j





a A / B = a A − aB = 0.2i + 4.461 j
B
30 m
30
o
60 m
r
θ
A
+θ
B
B
30 m
30o
α

vB

aB
60 m
+n
r
θ
+t

aA
β
β
A
30o

vA
+r
30 m
30
o
60 m
r
θ
A
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 14 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.

 
v A / B = v A − vB
y
C «Catch position»
2.1 m
x
y = yo + (vo )y t −
1 2
gt ⇒
2
2.1 = 0.9 + (30 sin 30 )t −
1
(9.81)t 2
2
⇒ t1, 2 =
Total flight time
2.98 s
0.08 s
1
= 2.73 s (the duration that fielder B must run)
4
x = (30 cos 30 )(2.98) = 77.3 m
Range: x = xo + (vo )x t ⇒
2.98 −
77.3 − 65 = 12.3 m (Fielder B must run 12.3 m in 2.73 seconds.)


12.3
= 4.52 m / s (cst ) ⇒
vB = 4.52i
Velocity of fielder B: vB =
2.73
Velocity components of the ball at C:
v x = (vo )x = 30 cos 30 = 25.98 m / s
in y-direction: v y = (vo )y − gt = 30 sin 30 − 9.81(2.98) = −14.19 m / s
in x-direction:



v A = 25.98i − 14.19 j
Velocity of the ball relative fielder




 
v A / B = v A − vB = (25.98i − 14.19 j ) − (4.52i )
⇒



v A / B = 21.46i − 14.19 j
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


v
wheel B ( vB ), its relative velocity with respect to A ( B / A ), its absolute


a
acceleration (a B ) and its relative acceleration with respect to A ( B / A ).
A
500 mm

vA
60°
800 mm
B
500 mm
In Cartesian Coordinates


v A = 1.5i



vB = vB cos 60i − vB sin 60 j



vB = 0.5vB i − 0.866vB j



vB = v A + vB / A


 
0.5vB i − 0.866vB j = 1.5i + vB / A

vA
A
A
800 mm
B

vB
500 mm
θ
x
60°
In Polar Coordinates

vA
y
120°
60°
β
800 mm
B
θ 30°
+θ
β
θ

vB
+r
Velocity:

vA
A
500 mm
θ
800 mm
120°
r=0.8 m
500
800
=
sin β sin 120
60°
β
+θ
β = 32.75
B
θ
30°
β
θ

vB
+r



v A = v A cos θer − v A sin θeθ



v A = 1.333er − 0.687eθ





v B = v B cos β er + v B sin β eθ = 0.841v B er + 0.541v B eθ





v B / A = v B − v A = vr er + vθ eθ
vr = r = 0
( r = 800 mm − constant
r = r = 0 )






vr er + vθ eθ = (0.841v B er + 0.541v B eθ ) − (1.333er − 0.687eθ )

0
θ = 27.25

er

eθ
→
0 = 0.841vB − 1.333
→
vθ = 0.541vB + 0.687
1.544
= 1.93 rad / s
vB = 1.585 m / s
vθ = 1.544 m / s
θ =
0.8


v B / A = 1.544eθ





v B / A = −1.544 sin θi − 1.544 cos θj = −0.707i − 1.373 j
A

vA
500 mm
θ
800 mm
120°
60°
β
+θ
B
θ 30°
β
θ

vB
+r
Acceleration:



a A = a A cos θer − a A sin θeθ
a A = 0.5 m / s 2



a A = 0.444er − 0.23eθ





a B = a B cos β er + a B sin β eθ = 0.841a B er + 0.541a B eθ





a B / A = a B − a A = ar er + aθ eθ
ar = r − rθ 2 = 0 − 0.8(1.93)2 = −2.98 m / s 2






ar er + aθ eθ = (0.841a B er + 0.541a B eθ ) − (0.444er − 0.23eθ )
A

aA
500 mm
θ
800 mm
120°
60°
β
+θ
B
θ 30° β
θ

aB
+r






ar er + aθ eθ = (0.841a B er + 0.541a B eθ ) − (0.444er − 0.23eθ )

er

eθ
→
ar = 0.841a B − 0.444

− 2.98
→
aθ = 0.541a B − 0.23
a B = −3.015 m / s 2
aθ = −1.41 m / s 2



a B / A = −2.98er − 1.4eθ