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Chapter 3: Vectors & Two-Dimensional Motion
Vectors and Their Properties All the physical quantities we will learn are classified either as a vector or scalar quantity. A scalar is specified by its magnitude, while a vector by its magnitude AND its direction.
Examples• Scalar : temperature, speed, mass, volume, length, etc.• Vector : displacement, velocity, acceleration, force, etc.
Vectors
Vectors and Their Properties
• Notation AAA
magnitude ;
• Equality of two vectors
BA
if their magnitudes and the directions are the same.
• Addition of vectors (geometrical method)
Commutative law of addition: ABBA
Vectors
Vectors and Their Properties
• Negative of a vector A
A
is defined as the vector that gives zero when added to A
A
has the same magnitude but opposite direction of A
• Subtraction of vectors
)( BABA
Vectors
Vectors and Their Properties
• Example 3.1: Taking a trip A car travels 20 km due northand then 35.0 km in a direction600 west of north.
What is the net effect of the car’strip?
39
km, 35.0km 20.0km 48
km 0.35km, 0.20
,
R
BA
RBA
Components of a vector
Components of a Vector
• A vector in two-dimension can be specified by a pair of coordinates
yxyx AAAAA
),(
xA
yA
x-component
y-component
x
y
x
y
yx
yx
AA
AA
AAA
AAAA
1
22
tan,tan
sin,cos
tan-1defined in (-900,900).Add 1800 when the vector is in 2nd or 3rd quadrant.
Components of a vector in different coordinates• A vector in two-dimension can be specified by a pair of coordinates
Bx
By
),(),( '' yxyx BBBBB
If a different coordinate system used, the components aredifferent to represent the samevector.
x
y
Vector addition by components:
),(
),(),(
yyxx
yxyx
BABA
BBAABA
But use the same coordinatesystem for both vectors
Components of a Vector
An example• Example 3.3: Take a hike
1st day : 25.0 km southeast2nd day : 40.0 km in a direction 60.0o north of east
A
B
Determine the components of the hiker’s displacements in the 1st and 2nd days.
km -17.7km)(0.707) 0.25(0.45sin
km 17.7km)(0.707) 0.25(0.45cos
AAAA
y
x
km 34.6km)(0.866) 0.40(0.60sinkm 20.0km)(0.500) 0.40(0.60cos
BBBB
y
x
Components of a Vector
1.24)/(tan km, 3.41
km 9.16km, 7.37122
xyyx
yyyxxx
RRRRR
BARBAR
Displacement, Velocity, & Acceleration in Two-Dimension
Displacement in 2D
• A position vector describes the position of an object at a time.
ii tr at timeector position v :
ff tr at timeector position v :
• An object’s displacement from ti to tf is defined by:
m :units SI if rrr
Displacement, Velocity, & Acceleration in Two-Dimension
Velocity in 2D
• An object’s average velocity during a time interval t is:
m/s :units SI trvav
• An object’s instantaneous velocity is:
m/s :units SI lim 0 trv t
Displacement, Velocity, & Acceleration in Two-Dimension
Acceleration in 2D
• An object’s average acceleration during a time interval t is:
2m/s :units SI tvaav
• An object’s instantaneous acceleration is:
20 m/s :units SI limtva t
Motion in Two-Dimension
Motion in 2D: horizontal and vertical direction
• In this chapter, we will learn movement of an object in both the x- and y-direction simultaneously under constant acceleration.
• An example: projectile motion under influence of gravity
Motion in Two-Dimension
Projectile motion under influence of gravity
• Let’s examine the motion of an object that is tossed into air but let’s neglect the effects of air resistance and the rotation of Earth.
• It was experimentally proven that the horizontal and vertical motions are completely independent of each other.
Motion in one direction has no effect on motion in the otherdirection.
So, in general, the equations of constant acceleration welearned in Lecture 2 follow separately for both the x-directionand the y-direction.
Motion in Two-Dimension
Projectile motion under influence of gravity
• Let’s assume that at t=0, the projectile leaves the origin with an initial velocity with an angle with the horizontal.0v
0
20
20
2
02
0
00
00000
221
,0;sin,cos
xxxx
xxx
xxxx
yxyx
vxavv
tvtatvx
vtavv
gaavvvv
x-direction:
Motion in Two-Dimension
Projectile motion under influence of gravity• Let’s assume that at t=0, the projectile leaves the origin with an initial velocity with an angle with the horizontal.0v
0
ygv
yavv
gttv
tatvy
gtvtavv
gaavvvv
y
yyy
y
yy
yyyy
yxyx
2
22121
,0;sin,cos
20
20
2
20
20
00
00000
y-direction:
Motion in Two-Dimension
Projectile motion under influence of gravity• Plug-in all the known quantities.
Equations that describe the motion in the y-direction:
Equations that describe the motion in the x-direction:
tvtvxvvv
x
xx
)cos(cosntant cos
000
000
ygvv
gttvy
gtvv
y
y
2)sin(21)sin(
sin
200
2
200
00
Velocity :
x
yyx v
vvvv 122 tan,
Motion in Two-Dimension
Projectile motion under influence of gravity• Trajectories as a function of the projection angle
Note : Given the displacement in x there are two corresponding projection angles.
o
o
o
o
o
Motion in Two-Dimension
Examples• Problem 3.5: Stranded explorers(a) What is the range of the package?
range of package
200 2
1 gttvyyy y
m 100)m/s 90.4( 22 ty
s 52.4t
tvxxx x00
m 181s) m/s)(4.52 0.40( x
s 52.4 m/s, 0.40m, 00.0 00 tvx x
Motion in Two-Dimension
Examples• Problem 3.5: Stranded explorers(b) What are the velocity components of the package at impact?
range of package
m/s 40.0m/s)cos0 0.40(cos0
vvx
m/s 3.44s) m/s)(4.52 80.9(00.0
sin0
gtvvy
x component:
y component:
Motion in Two-Dimension
Examples• Problem 3.6: The long jump(a) How long does it take for the jumper to reach the max. height?
s 384.0m/s 80.9
)20.0m/s)(sin 0.11(
sin
0sin
2
o
00max
max00
gvt
gtvvy
y component:
=20.0ov 0
= 11.0 m/s
at max.height vy =0
Motion in Two-Dimension
Examples• Problem 3.6: The long jump(b) Find the maximum height he reaches.
m 722.0s) 384.0)(m/s 80.9)(2/1(
s) 384.0)(0.20sinm/s)( 0.11()()2/1()sin(
22
2maxmax00max
tgtvy
y component:
=20.0ov 0
= 11.0 m/s
From part (a) s 384.0max t
m 7.94s) 768.0)(0.20cosm/s)( 0.11(
)cos(
s 0.768s) 384.0(22
00
max
tvx
tt
Motion in Two-Dimension
Examples• Problem 3.6: The long jump(c) Find the horizontal distance he jumps.
Displacement in x:
=20.0ov 0
= 11.0 m/s
s 384.0max t
m/s 386s) 3.14)(m/s 0.20(
m/s 1000.12
2
0
tavv xxx
Motion in Two-Dimension
Examples• Problem 3.8: The rocket
Eq.3.14c:
m/s 1040.1
m) 1000.1)(m/s 08.9(20
2
2
322
20
2
y
y
yy
v
v
ygvv
(b) Find the rocket’s velocity in x direction. Eq.3.12a:
s 3.14)m/s 80.9(0m/s 1040.1 22
0
tt
tavv yyy
Eq.3.11a:
(a) Find the rocket’s velocity in y direction.
Motion in Two-Dimension
Examples• Problem 3.8: The rocket(c) Find the magnitude and direction of the rocket’s velocity.
9.19m/s 386
m/s 1040.1tantan
m/s 411m/s) 386(m/s) 1040.1(
211-
222
22
x
y
yx
vv
vvv
Relative Velocity
What is relative velocity?
• Relative velocity relates velocities measured by two different observers, one moving with respect to the other.
•The measured velocity of an object depends on the velocity of the observer with respect to the object.
• Measurements of velocity depend on the reference frame of the observer where the reference frame is a just coordinate system used to measure physical quantities such as velocity, acceleration etc. Most of time, we will use a stationary frame of reference, relative to earth, but occasionally we will use a moving frame of reference.
Relative Velocity
What is relative velocity? (cont’d)• More elaborate definition: - Let’s define E as a stationary observer with respect to Earth - A and B as two moving cars
BCar in observer an by measured asA Car ofposition theEby measured as BCar ofposition theEby measured asA Car ofposition the
AB
BE
AE
rrr
)()()()()()(
222
111
trtrtrtrtrtr
BEAEAB
BEAEAB
BEAEAB
ABABAB
rrrtrtrr
etc. )()( 12
Relative Velocity
What is relative velocity? (cont’d)
BCar in observer anby measured asA Car of velocity theEby measured as BCar of velocity theEby measured asA Car of velocity the
AB
BE
AE
vvv
tr
tr
tr BEAEAB
BEAEAB vvv
12 ttt BEAEAB rrr
relative to; with respect to
tr
tr
tr BE
t
AE
t
AB
t
000limlimlim
Examples• Example 3.10: Crossing a river
Relative Velocity
determined be toEarth, to relativeboat theofvelocity :
km/h 10.0 water, the torelativeboat theofvelocity :
km/h 5.00 Earth, torelative flowriver theof velocity :
BE
BR
RE
v
v
v
REBRBE vvv
Examples
Relative Velocity
),( REyBRyRExBRxREBRBE vvvvvvv
km/h) km/h,0 00.5(),(
)km/h 0.10,km/h 0(),(
REyRExRE
BRyBRxBR
vvv
vvv
km/h) 10.0 km/h, 00.5()km/h 00.00.10,km/h 00.500.0(
BEv
6.26tan
km/h 2.11
1
22
BEy
BEx
BEyBExBE
vv
vvv
• Example 3.10: Crossing a river (cont’d)