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Chapter 2
Motion Along a Straight Line
Linear motion
In this chapter we will consider moving objects:
• Along a straight line
• With every portion of an object moving in the same direction and at the same rate (particle-like motion)
Types of physical quantities
• In physics, quantities can be divided into such general categories as scalars, vectors, matrices, etc.
• Scalars – physical quantities that can be described by their value (magnitude) only
• Vectors – physical quantities that can be described by their value and direction
Distance, position, and displacement
• Distance (scalar) a total length of the path traveled regardless of direction (SI unit: m)
• In each instance we choose an origin – a reference point, convenient for further calculations
• Position of an object (vector) is described by the shortest distance from the origin and direction relative to the origin
• Displacement (vector) – a change from position x1 to position x2
12 xxx
Velocity and speed
• Average speed (scalar) - a ratio of distance traveled (over a time interval) to that time interval (SI unit: m/s)
• Average velocity (vector) - a ratio of displacement (over a time interval) to that time interval
• Instantaneous velocity (vector) – velocity at a given instant
• Instantaneous speed (scalar) – a magnitude of an instantaneous velocity
t
xvavg
t
xv
t
0
lim
12
12
tt
xx
dt
dx
Acceleration
• Average acceleration (vector) - a ratio of change of velocity (over a time interval) to that time interval (SI unit = (m/s)/s = m/s2)
• Instantaneous acceleration (vector) – a rate of change of velocity at a given instant
2
2
dt
xd
t
vaavg
12
12
tt
vv
t
va
t
0
limdt
dv
dt
dx
dt
d
Case of constant acceleration
• Average and instantaneous accelerations are the same
• Conventional notation
• Then tt 2
01 xx 01 txx 2
01 vv vv 2
t
vaa avg
12
12
tt
vv
0
0
t
vv
atvv 0
Case of constant acceleration
• Average and instantaneous accelerations are the same
• Conventional notation
• Then
t
xvavg
00
t
xx tvxx avg 0
22021 vvvv
vavg
20
atv
12
12
tt
xx
2
)( 00 atvv
2
2
00
attvxx
tt 2
01 xx 01 txx 2
01 vv vv 2
Case of constant acceleration
dt
dxv
dt
dva
atvv 0
2
2
00
attvxx
Case of constant acceleration
Case of constant acceleration
To help you solve problems
Equations Missing variables
2
2
00
attvxx
atvv 0
)(2 02
02 xxavv
2
)( 00
tvvxx
2
2
0
atvtxx
v
0xx
t
a
0v
Alternative derivation
Using definitions and initial conditions
we obtain
dt
dva
2
2
00
attvxx
dtadv dt
dxv vdtdx
Case of free-fall acceleration
• At sea level of Earth’s mid-latitudes all objects fall (in vacuum) with constant (downward) acceleration of
a = - g ≈ - 9.8 m/s2 ≈ - 32 ft/s2
• Conventionally, free fall is along a vertical (upward) y-axis
gtvv 0
2
2
00
gttvyy
Graphical representation
Graphical representation
Graphical representation
Graphical representation
Graphical representation
Graphical representation
Graphical representation
Graphical integration
1
0
)(01
t
t
dttavv
1
0
)(01
t
t
dttvxx
dt
dxv
dt
dva
Answers to the even-numbered problems
Chapter 2:
Problem 12: (a) – 6 m/s.(b) negative x direction (c) 6 m/s(e) Yes(f) No
Answers to the even-numbered problems
Chapter 2:
Problem 16: 5.9 m
Answers to the even-numbered problems
Chapter 2:
Problem 28: (a) 2.5 s.
Answers to the even-numbered problems
Chapter 2:
Problem 42: (a) 3.70 m/s.(b) 1.74 m/s (c) 0.154 m
Answers to the even-numbered problems
Chapter 2:
Problem 68: (a) 5.00 m/s(b) 1.67 m/s2
(c) - 7.5 m(d) - 3 s