12–74.
The velocity of a particle is v = 53i + (6 - 2t)j6 m>s, where t
is in seconds. If r = 0 when t = 0, determine the
displacement of the particle during the time interval
t = 1 s to t = 3 s.
SOLUTION
Position: The position r of the particle can be determined by integrating the
kinematic equation dr = vdt using the initial condition r = 0 at t = 0 as the
integration limit. Thus,
dr = vdt
t
r
L0
dr =
L0
C 3i + (6 - 2t)j D dt
r = c 3ti + A 6t - t2 B j d m
When t = 1 s and 3 s,
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
r t = 1 s = 3(1)i + C 6(1) - 12 D j = [3i + 5j] m>s
r t = 3 s = 3(3)i + C 6(3) - 32 D j = [9i + 9j] m>s
Thus, the displacement of the particle is
¢r = r t = 3 s - r t = 1 s
= (9i + 9j) - (3i + 5j)
= {6i + 4j} m
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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12–79.
A car traveling along the straight portions of the road has
the velocities indicated in the figure when it arrives at
points A, B, and C. If it takes 3 s to go from A to B, and then
5 s to go from B to C, determine the average acceleration
between points A and B and between points A and C.
y
vC
x
vB
B
30 m/s
40 m/s
C
45
SOLUTION
vA = 20 i
vB = 21.21 i + 21.21j
vA
20 m/s
A
vC = 40i
21.21 i + 21.21 j - 20 i
¢v
=
¢t
3
aAB = { 0.404 i + 7.07 j } m>s2
aAC =
40 i - 20 i
¢v
=
¢t
8
aAC = { 2.50 i } m>s2
Ans.
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
aAB =
Ans.
A
ns
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–87.
Pegs A and B are restricted to move in the elliptical slots
due to the motion of the slotted link. If the link moves with
a constant speed of 10 m/s, determine the magnitude of the
velocity and acceleration of peg A when x = 1 m.
y
A
D
C
SOLUTION
Velocity: The x and y components of the peg’s velocity can be related by taking the
first time derivative of the path’s equation.
x2
+ y2 = 1
4
1
#
#
(2xx) + 2yy = 0
4
1 #
#
xx + 2yy = 0
2
v
x
10 m/s
B
x2
4
y2
1
or
1
xv + 2yvy = 0
2 x
(1)2
+ y2 = 1
4
th an Th
sa eir d i is w
s
le
o p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
At x = 1 m,
(1)
y =
23
m
2
Here, vx = 10 m>s and x = 1. Substituting these values into Eq. (1
(1),
1),
23
1
(1)(10) + 2 ¢
≤ vy = 0
2
2
vy = -2.887
7m
m>s
>s = 2.887
2.887 m
m>s
>s T
Thus, the magnitude of the peg’s velocity is
887
72 = 10
110.4
.4 m>s
m
v = 2vx 2 + vy 2 = 2102 + 2.887
Ans.
Acceleration: The x and y componentss of
peg’ss ac
acceleration
can be related by
of tthe peg
ceele
taking the second time derivative of the pa
path’s
h’s eequation.
quation.
1 # #
##
##
# #
y) = 0
(xx + xx) + 2(yy + yy)
2
1 #2
##
##
#
A x + xx B + 2 A y 2 + yy B = 0
2
or
1
A v 2 + xax B + 2 A vy 2 + yay B = 0
2 x
Since vx is constant, ax = 0. When x = 1 m, y =
(2)
23
m, vx = 10 m>s, and
2
vy = -2.887 m>s. Substituting these values into Eq. (2),
23
1
a d = 0
A 102 + 0 B + 2 c ( - 2.887)2 +
2
2 y
ay = - 38.49 m>s2 = 38.49 m>s2 T
Thus, the magnitude of the peg’s acceleration is
a = 2ax 2 + ay 2 = 202 + ( -38.49)2 = 38.5 m>s2
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–101.
It is observed that the skier leaves the ramp A at an angle
uA = 25° with the horizontal. If he strikes the ground at B,
determine his initial speed vA and the speed at which he
strikes the ground.
vA
uA
A
4m
3
5
4
100 m
SOLUTION
Coordinate System: x - y coordinate system will be set with its origin to coincide
with point A as shown in Fig. a.
4
x-motion: Here, xA = 0, xB = 100 a b = 80 m and (vA)x = vA cos 25°.
5
xB = xA + (vA)xt
80 = 0 + (vA cos 25°)t
80
t =
vA cos 25°
(1)
th an
sa eir d i w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
+ B
A:
3
y-motion: Here, yA = 0, yB = - [4 + 100 a b ] = - 64 m and (vA)y = vA ssin
in 25°
in
25
5
and ay = - g = - 9.81 m>s2.
A+ c B
yB = yA + (vA)y t +
1
a t2
2 y
- 64 = 0 + vA sin 25° t +
1
( - 9.81)t2
2
4.905t2 - vA sin 25° t = 64
Substitute Eq. (1) into (2) yieldS
4.905 ¢
¢
(2)
2
80
8800
n 25° ¢
≤ = vA sin
≤ = 64
vA cos 25°
yA co
c s 25°
cos
2
80
≤ = 20.65
vA cos 25°
80
= 4.545
vA cos 25°
vA = 19.42 m>s = 19.4 m>s
Ans.
Substitute this result into Eq. (1),
t =
80
= 4.54465
19.42 cos 25°
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
B
12–101. continued
Using this result,
A+ c B
(vB)y = (vA)y + ay t
= 19.42 sin 25° + ( -9.81)(4.5446)
= - 36.37 m>s = 36.37 m>s T
And
+ B
A:
(vB)x = (vA)x = vA cos 25° = 19.42 cos 25° = 17.60 m>s :
Thus,
vB = 2(vB)2x + (vB)2y
= 236.372 + 17.602
Ans.
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
= 40.4 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–112.
The baseball player A hits the baseball at vA = 40 ft>s and
uA = 60° from the horizontal. When the ball is directly
overhead of player B he begins to run under it. Determine
the constant speed at which B must run and the distance d
in order to make the catch at the same elevation at which
the ball was hit.
vA = 40 ft/s
A
θA
15 ft
B
C
vA
d
SOLUTION
Vertical Motion: The vertical component of initial velocity for the football is
(v0)y = 40 sin 60° = 34.64 ft>s. The initial and final vertical positions are (s0)y = 0
and sy = 0, respectively.
( + c ) sy = (s0)y + (v0)y t +
0 = 0 + 34.64t +
1
(a ) t2
2 cy
1
( - 32.2) t2
2
t = 2.152 s
+ )
(:
sx = (s0)x + (v0)x t
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in
is
t la
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
Horizontal Motion: The horizontal component of velocity for the baseball
ll is
ns are
are
(v0)x = 40 cos 60° = 20.0 ft>s. The initial and final horizontal positions
(s0)x = 0 and sx = R, respectively.
R = 0 + 20.0(2.152) = 43.03 ft
The distance for which player B must travel in order to catch
baseball
tcch tthe
eb
aseball is
d = R - 15 = 43.03 - 15 = 28.0 ft
Ans.
Player B is required to run at a same speed as the h
horizontal
component
of velocity
orrizonta com
mpon
of the baseball in order to catch it.
vB = 40 cos 60° = 20.0 ft s
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.