TUTORIAL+CHAPTER+1 Solution
TUTORIAL+CHAPTER+1 Solution
TUTORIAL+CHAPTER+1 Solution
Solution Q1
Kinematics:
and
(
()
Answer
(
()
Answer
Solution Q2
(a) Determine the position and the velocity of the particle when t=14 s
Split into three parts: A, B and C
Part A ( 0<t<6) constant deceleration
When t=6s, 8 V
A
= m/s
When t=0s; t a V V
A A A
+ =
O
) (
) )( ( ) ( 6 4 V 8
A o
+ =
16 V
A
=
O
) ( m/s
When 0 V
A
= , 4 t = s
When t=0s, s=0
CHAPTER 1: KINEMATICS OF PARTICLE
When t=6s; ( ) ( )
2
A A A o A
t a
2
1
t V S S + + =
O
) )( ( ) (
2
A
6 4
2
1
6 16 0 S + + =
24 S
A
= m
When t=4s; ( ) 32 S
A 4 t
=
=
m
Part B ( 6<t<10) zero acceleration, a =0
When t=6s, ( ) 8 V V
B A
= =
O
m/s
When t=10s, 8 V
B
= m/s
When t=6s, ( ) 24 S S
B A
= =
O
m
When t=10s, ( ) ( )
2
B B B o B
t a
2
1
t V S S + + =
O
56 S
B
= m
Part C ( 10<t<14) constant acceleration
When t=10s, ( ) 8 V V
C B
= =
O
m/s
When t=14s, t a V V
C C C
+ =
O
) (
) )( ( 4 4 8 V
C
+ =
8 V
C
= m/s Answer
When 0 V
C
= , 12 t = s
When t=10s, ( ) 56 S S
C B
= =
O
m/s
When t=14s, ( )
2
C C C C
t a
2
1
t V S S + + =
O O
) (
56 S
C
= m Answer
When t=12s; ( ) 64 S
C 12 t
=
=
m
(b) Determine the total distance traveled by the particle when t=14 s
Total distance = 32+(32-24)+24+56+(64-56)+(64-56)=136 m Answer
CHAPTER 1: KINEMATICS OF PARTICLE
Solution Q3
graph: For the time interval , the condition is when
(
When
(
For the interval , the initial condition is
when .
(
( )
Thus, when
Answer
Also, the change in velocity is equal to the area under the graph. Thus
()() (
The graph is shown as
CHAPTER 1: KINEMATICS OF PARTICLE
graph: For the time interval , the initial condition is when
(
)
When ,
(
)
For the time interval
( )
)
When
,
()
()
CHAPTER 1: KINEMATICS OF PARTICLE
The graph is shown as
Solution Q4
Horizontal motion: ( ) 228 0 5 t 30 v t v x . cos cos = = =
O O
o
=
O
30 v
772 4
t
cos
.
Vertical motion: ( ) ( ) ( )
2 2
t 81 9
2
1
t 30 v 2 at
2
1
t v y y . sin sin + = + + =
O O O
o
( )
( )
2
t 81 9
2
1 772 4
2 048 3 .
cos
sin .
. + =
o
o
s 59 0 t . =
Therefore; 34 9 v . =
O
m/s Answer
Solution Q5
Coordinate system: The coordinate system will be set so that its origin coincides with
point A.
-motion: Here, (
and . Thus
(
CHAPTER 1: KINEMATICS OF PARTICLE
()
(1)
-motion: Here, (
and
. Thus
()
()
()(
(2)
Solving Eqs. (1) and (2) yields
Answer
Solution Q6
(a) At point A
Given 50 v
A
= m/s
89 8 25 81 9 25 a a
N
. cos . cos = = = m/s
2
Then 89 8
r
50
r
v
a
2 2
N
. = = =
281 r = m Answer
(a) At highest point 0 v
y
= and v 25 v v
A x
= = cos
Then 81 9 a
N
. = m/s
2
CHAPTER 1: KINEMATICS OF PARTICLE
Therefore
( )
81 9
r
25 50
r
v
a
2 2
N
.
cos
=
= =
209 r = m Answer
Solution Q7
, therefore
( )
,
( )
( )
|
( )
For
()
( )
Answer
( )
Answer
()
()
()
()
Answer
CHAPTER 1: KINEMATICS OF PARTICLE
Solution Q8
Position coordinates: By referring to Fig. (a), the length of the two ropes written in terms of
the position coordinates
and
are
(1)
and
(2)
Eliminating
(3)
Time derivative: Taking the time derivative of Eq. (3)
()
Here,
. Hence,
Answer
Solution Q9
Take positive if downward.
Constraint of cable on left: constant = +
B A
x 3 x 2
CHAPTER 1: KINEMATICS OF PARTICLE
0 = +
B A
v 3 v 2
A B
v
3
2
v =
Then
A B
a
3
2
a =
Constraint of cable on right: constant = +
C B
x 2 x
0 = +
C B
v 2 v
Hence
B c
v
2
1
v =
Then
B C
a
2
1
a =
Constraint of point D on cable:
constant = +
D A
x x 2
Hence
B A D
v 3 v 2 v = =
B D
a 3 a =
(a) Given 200 a
B
=
mm/s
2
( ) 300 200
2
3
a
A
= = mm/s
2
@ ( ) |
2
300mm/s
( ) 100 200
2
1
a
c
= = mm/s
2
@ ( ) |
2
100mm/s
( ) 600 200 3 a
D
= = mm/s
2
@ ( ) +
2
600mm/s
(b) Given ( ) ( ) ( ) 0 v v v
D C A
= = =
o o o
When t = 10 s; ( ) t a v v
A A A
+ =
( )( ) 3000 10 300 0 v
A
= + = mm/s @ ( ) | mm/s 3000
( )( ) 2610 10 200 610 v
B
= + = mm/s @ ( ) + mm/s 2610
( )( ) 1000 10 100 0 v
C
= + = mm/s @ ( ) | mm/s 1000
( )( ) 6000 10 600 0 v
D
= + = mm/s @ ( ) + mm/s 6000
(c) Relative velocity D to B: ( ) + = = = mm/s
/
3390 2610 6000 v v v
B D B D
CHAPTER 1: KINEMATICS OF PARTICLE
Solution Q10
A B A B
v v v
/
+ =
A B A B
v v v =
/
From x-axis:
( ) ( ) ( )
x A x B x A B
v v v =
/
( ) 30 43 0 30 50 v
x A B
. cos
/
= = km/h
From y-axis:
( ) ( ) ( )
y A y B y A B
v v v =
/
( ) ( ) 0 65 40 30 50 v
y A B
. sin
/
= = km/h
Magnitude 1 78 v
A B
.
/
= km/h Answer
Angle = 3 56
A B
.
/
u
Answer
Solution Q11
Relative Velocity:
Equating and component, we have
(1)
(2)
Solving Eqs. (1) and (2) yields
CHAPTER 1: KINEMATICS OF PARTICLE
Answer
Thus, the time required by the boat to travel from point A to B is
Answer