Problem (Q1):
Evaluate each of the following to three significant figures and express each answer in SI units: (a) (0.631
Mm)/(8.60 kg)
2(b) (35 mm)
2*(48 kg)
3SOLUTION
(
)
( )
(
)
( )
6 2 2 2 2 3 2 20.631 10
m
8532 m
(a)
0.631 Mm / 8.60 kg
kg
8.60
kg
8.53 10
m / kg
8.53 km / kg
=
=
=
=
(
) (
2)
2( )
2(
)
2 3 2 3(b)
35 mm
48 kg
=
35 10
−m
48 kg
=
135 m / kg
A sphere is fired downwards into a medium with an initial
speed of
. If it experiences a deceleration of
where t is in seconds, determine the
distance traveled before it stops.
a
= (-6t) m
>s
2,
27 m
>s
SOLUTION
Velocity:
at
. Applying Eq. 12–2, we have
(1)
At
, from Eq. (1)
Distance Traveled:
at
. Using the result
and applying
Eq. 12–1, we have
(2)
At
, from Eq. (2)
A
ns.
s
= 27(3.00) - 3.00
3= 54.0 m
t
= 3.00 s
s
=
A
27t - t
3B
m
L
s 0ds
=
L
t 0A
27 - 3t
2B
dt
A
+ T
B
ds
= vdt
v
= 27 - 3t
2t
0= 0 s
s
0= 0 m
0 = 27 - 3t
2t
= 3.00 s
v
= 0
v
=
A
27 - 3t
2B
m
>s
L
v 27dv
=
L
t 0-6tdt
A
+ T
B
dv
= adt
t
0= 0 s
v
0= 27 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.
Th
is
w
ork
w
ork
54.0
is
54.0
pro
te
ct
ed
b
y
U
ni
te
d
St
at
es
co
pyri
gh
t
and applying
la
w
s
and applying
a
nd
is
pro
vi
de
d
pro
vi
de
d
54.0 m
so
le
ly
fo
r t
he
u
se
o
f i
nst
ru
ct
ors
in
te
ach
in
g
te
ach
in
g
and applying
th
ei
r co
urse
s
an
d
an
d
asse
ssi
ng
st
ud
en
t l
ea
rn
in
g.
D
isse
mi
na
tio
n
D
isse
mi
na
tio
n
or
sa
le
o
f a
ny
pa
rt
of
th
th
is
is
w
ork
(in
cl
ud
in
g
on
th
e
W
orl
d
W
id
e
W
id
e
W
eb
)
w
ill
d
est
ro
y
th
e
in
te
gri
ty
in
te
gri
ty
of
th
e
w
ork
an
d
is
no
t p
ermi
tte
d.
pe
rmi
tte
d.
12–27.
A particle is moving along a straight line such that when it
is at the origin it has a velocity of
If it begins to
decelerate at the rate of
where v is in
determine the distance it travels before it stops.
m
>s,
a
=
1-1.5v
1>22 m>s
2,
4 m
>s.
SOLUTION
(1)
(2)
From Eq. (1), the particle will stop when
A
ns.
s
|
t= 2.667= 4(2.667) - 1.5(2.667)
2+ 0.1875(2.667)
3= 3.56 m
t
= 2.667 s
0 = (2 - 0.75t)
2s
= 4t - 1.5t
2+ 0.1875t
3L
s 0ds
=
L
t 0(2 - 0.75t)
2dt
=
L
t 0(4 - 3t + 0.5625t
2) dt
v
= (2 - 0.75t)
2m
>s
2
av
12- 2
b = -1.5t
2v
12 v 4= - 1.5t
t 0L
v 4v
-12dv
=
L
t 0- 1.5 dt
a
=
dv
dt
= - 1.5v
1 2© 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.
Th
is
w
ork
is
pro
te
ct
ed
pro
te
ct
ed
= 3.56
by
U
ni
te
d
St
at
es
co
pyri
gh
t l
aw
s
a
nd
is
pro
vi
de
d
so
le
ly
so
le
ly
3.56
fo
r
3.56 m
th
e
use
o
f i
nst
ru
ct
ors
in
te
ach
in
g
th
ei
r co
urse
s
an
d
asse
ssi
ng
asse
ssi
ng
m
st
ud
en
t l
ea
rn
in
g.
D
isse
mi
na
tio
n
(2)
or
sa
le
o
f a
ny
pa
rt
of
th
is
w
ork
w
ork
(in
cl
ud
in
g
on
th
e
W
orl
d
W
id
e
W
eb
)
w
ill
d
est
ro
y
th
e
in
te
gri
ty
of
of
th
e
th
e
w
ork
an
d
is
no
t p
ermi
tte
d.
A particle travels to the right along a straight line with a
velocity
where s is in meters.
Determine its deceleration when s = 2 m.
v
= [5
>14 + s2] m>s,
SOLUTION
When
A
ns.
a
= -0.116 m
>s
2s
= 2 m
a
=
- 25
(4 + s)
35
(4 + s)
a
- 5 ds
(4 + s)
2b = a ds
dv
=
-5 ds
(4 + s)
2v dv
= a ds
v
=
5
4 + 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.
Th
is
w
ork
is
pro
te
ct
ed
b
y
by
U
ni
te
d
St
at
es
co
pyri
gh
t l
aw
s
a
nd
is
pro
vi
de
d
so
le
ly
fo
r t
he
th
e
use
o
f i
nst
ru
ct
ors
in
te
ach
in
g
th
ei
r co
urse
s
an
d
asse
ssi
ng
st
ud
en
t
st
ud
en
t l
ea
rn
in
g.
D
isse
mi
na
tio
n
A
ns
or
sa
le
o
f a
ny
pa
rt
of
th
is
w
ork
(in
cl
ud
in
g
(in
cl
ud
in
g
on
th
e
W
orl
d
W
id
e
W
eb
)
w
ill
d
est
ro
y
th
e
in
te
gri
ty
of
th
e
w
ork
w
ork
an
d
is
no
t p
ermi
tte
d.
12–30.
As a train accelerates uniformly it passes successive
kilometer marks while traveling at velocities of
and
then
. Determine the train’s velocity when it passes
the next kilometer mark and the time it takes to travel the
2-km distance.
10
m
>s
2 m
>s
SOLUTION
Kinematics:
For the first kilometer of the journey,
,
,
,
and
. Thus,
For the second kilometer,
,
,
,
and
. Thus,
A
ns.
For the whole journey,
,
, and
. Thus,
A
ns.
t
= 250 s
14 = 2 + 0.048t
A
:
+
B
v
= v
0+ a
ct
0.048 m
>s
2v
= 14 m
>s
v
0= 2 m
>s
v
= 14 m
>s
v
2= 10
2+ 2(0.048)(2000 - 1000)
A
:
+
B
v
2= v
02+ 2a
c(s - s
0)
0.048 m
>s
2s
= 2000 m
s
0= 1000 m
v
0= 10 m
>s
a
c= 0.048 m
>s
210
2= 2
2+ 2a
c(1000 - 0)
A
:
+
B
v
2= v
02+ 2a
c(s - s
0)
s
= 1000 m
s
0= 0
v
= 10 m
>s
v
0= 2 m
>s
a
c=
a
c=
© 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.
