CHAPTER 2
Motion in One Dimension
KINEMATICS
• Kinematics describes motion while ignoring the external
agents that might have caused or modified the motion
• Motion represents a continual change in an object’s
position
• In this chapter, we consider only motion in one
dimension, and describe the moving object as a particle
regardless of its size (the particle model)
POSITION
• The motion of a particle is completely known if the
particle’s position in space is known at all times
• You need a coordinate system to measure the
position.
POSITION-TIME GRAPH
• The position-time graph
shows the motion of the
particle (car).
• This is a permanent record
of the motion.
• Q: when was the particle to
the right of the origin? to
the left? When was it at the
origin? When was it moving
towards left? right? When
was it at rest?
DISPLACEMENT
• Displacement is defined as the change in position during
some time interval
– Represented as x
x ≡ xf - xi
– SI units are meters (m)
– x can be positive or negative
• Different than distance
– Distance is the length of a path followed by a particle
DISTANCE VS. DISPLACEMENT
– AN EXAMPLE
• Assume a player moves from one end of the court to the
other and back
• Distance is twice the length of the court
– Distance is always positive
• Displacement is zero
Δx = xf – xi = 0 since xf = xi
• Q:Do you have different words for position and
displacement in your mother language?
DISTANCE VS. DISPLACEMENT
– AN EXAMPLE
Examples A particle moves . . .
o
From x = 5 m to x = 12 m:
o
∆x = 7 m (positive direction)
o
From x = 5 m to x = 1 m:
o
∆x = -4 m (negative direction)
o
From x = 5 m to x = 200 m to x = 5 m:
o
∆x = 0 m
VECTORS AND SCALARS
• Vector quantities need both magnitude (size or numerical
value) and direction to completely describe them
– Will use + and – signs to indicate vector directions in
this chapter
• Scalar quantities are completely described by magnitude
only
AVERAGE VELOCITY
• The average velocity is the rate at which the
displacement occurs
v x , avg 
x xf  xi

t
t
– The x indicates motion along the x-axis
• The dimensions are length / time [L/T]
• The SI units are m/s
• Is also the slope of the line in the position – time graph
AVERAGE SPEED
• Speed is a scalar quantity
– Has the same units as velocity
– Defined as total distance / total time:
v avg 
d
t
• The speed has no direction and is always expressed as a
positive number
• Neither average velocity nor average speed gives details
about the trip described
AVERAGE SPEED AND AVERAGE
VELOCİTY
• The average speed is not the magnitude of the average
velocity
– For example, a runner ends at her starting point
– Her displacement is zero
– Therefore, her average velocity is zero
– However, the distance traveled is not zero, so the
average speed is not zero
AVERAGE VELOCITY GRAPHICALLY
INSTANTANEOUS VELOCITY, EQUATIONS
• The general equation for instantaneous velocity is:
v x  lim
t 0
x dx

t
dt
• The instantaneous velocity can be positive, negative, or
zero
• The instantaneous velocity indicates what is happening
at every point of time
INSTANTANEOUS VELOCITY, GRAPH
• The instantaneous
velocity is the slope of the
line tangent to the x vs. t
curve
– This would be the
green line
• The light blue lines show
that as t gets smaller,
they approach the green
line
INSTANTANEOUS VELOCITY, GRAPH
• Another graph illustrating
the limiting process
graphically. As ∆𝑡 → 0 the
lines approach the
tangent at t0 .
• The instantaneous
velocity at t=2 s is the
slope of the tangent to the
graph there.
INSTANTANEOUS SPEED
• The instantaneous speed is the magnitude of the
instantaneous velocity. It is always a positive number.
• The instantaneous speed has no direction associated
with it
• “Velocity” and “speed” will indicate instantaneous values.
• Q: Do you have different words for speed and velocity in
your mother language?
ANALYSIS MODEL: A PARTICLE WITH
CONSTANT VELOCITY
• Constant velocity indicates the instantaneous velocity at
any instant during a time interval is the same as the
average velocity during that time interval
– vx = vx, avg
– The mathematical representation of this situation is the
equation
vx 
x xf  xi

t
t
or
xf  xi  v x t
– Common practice is to let ti = 0 and the equation
becomes: xf = xi + vx t (for constant vx)
PARTICLE UNDER CONSTANT VELOCITY,
GRAPH
• The graph represents the
motion of a particle under
constant velocity
• Q: Why is it a straight line?
• The slope of the graph is the
value of the constant velocity
• The y-intercept is xi , i.e. the
position at t=0
• Q: What does a negative
slope mean? a zero slope?
MODEL: A PARTICLE UNDER CONSTANT
SPEED
• A particle under constant velocity moves with a constant
speed along a straight line
• A particle can also move with a constant speed along a
curved path. This is NOT motion with constant velocity.
• This can be represented with a model of a particle
under constant speed
v
d
t
AVERAGE ACCELERATION
• Acceleration is the rate of change of the velocity (not the
speed). Average acceleration is defined as:
ax ,avg 
v x v xf  v xi

t
tf  t i
• Dimensions are L/T2
• SI units are m/s²
• In one dimension, positive and negative can be used to
indicate direction
INSTANTANEOUS ACCELERATION
• The instantaneous acceleration is the limit of the average
acceleration as t approaches 0
v x dv x d 2 x
ax  lim

 2
t 0 t
dt
dt
• The term acceleration will mean instantaneous
acceleration
– If average acceleration is wanted, the word average
will be included
INSTANTANEOUS ACCELERATION –
GRAPH
• The slope of the velocity-time
graph is the acceleration
• The green line represents the
instantaneous acceleration
• The blue line is the average
acceleration
GRAPHICAL COMPARISON
• (a): the position-time graph
• (b):The velocity-time graph is
found by measuring the slope of
the position-time graph at every
instant
• The acceleration-time graph is
found by measuring the slope of
the velocity-time graph at every
instant.
• Q:When was the particle
speeding up? Slowing down?
Moving with constant velocity?
NOTES ABOUT ACCELERATION
• When an object’s velocity and acceleration are in the same
direction, the object is speeding up
• When an object’s velocity and acceleration are in the opposite
direction, the object is slowing down
• Negative acceleration does NOT necessarily mean the object
is slowing down
• Positive acceleration does NOT necessarily mean the object is
slowing down
– If the acceleration and velocity have the same sign, the
object is speeding up, if they have different signs it is
slowing down
POSITION, VELOCITY AND ACCELERATION
USING CALCULUS
The position of a particle as a function of time is given by
the expression:
x(t )  5  6t  7t 2  2t 4 (m)
Then, we have, for example:
x(0)  x0  5m, x(1)  16m
The x-t graph of the motion over the interval t=0 to t=1.8 s is
shown below:
POSITION, VELOCITY AND ACCELERATION
USING CALCULUS
The velocity of the particle as a function of time is given by
dx
the expression:
v(t ) 
 6  14t  8t 3
dt
So, we have, for example:
v(0)  v0  6m / s, v(1)  12m / s
The v-t graph of the motion over the interval t=0 to t=1.8 s is
shown below:
POSITION, VELOCITY AND ACCELERATION
USING CALCULUS
The acceleration of the particle as a function of time is given
dx
by the expression:
a (t ) 
 14  24t 2
dt
MOTION DIAGRAMS
• A motion diagram can be formed by imagining the
stroboscope photograph of a moving object
KINEMATIC EQUATIONS
• The kinematic equations can be used with any particle
under uniform (constant) acceleration
• The kinematic equations may be used to solve any
problem involving one-dimensional motion with a
constant acceleration
• You may need to use two of the equations to solve one
problem
• Many times there is more than one way to solve a
problem
KINEMATIC EQUATIONS, 1
• For constant ax,
• 𝑎𝑎𝑣𝑔 = 𝑎𝑥 →→
v xf  v xi  ax t
• Can determine an object’s velocity at any time t when we
know its initial velocity and its acceleration
– Assumes ti = 0 and tf = t
• Does not give any information about displacement
KINEMATIC EQUATIONS, 2
• For constant acceleration,
v x,avg 
v xi  v xf
2
• The average velocity can be expressed as the arithmetic
mean of the initial and final velocities
– This applies only in situations where the acceleration
is constant
KINEMATIC EQUATIONS, 3
• For constant acceleration,
1
xf  xi  v x,avg t  xi  v xi  v fx  t
2
• This gives you the position of the particle in terms of time
and velocities
• Doesn’t give you the acceleration
KINEMATIC EQUATIONS, 4
• For constant acceleration,
1 2
xf  xi  v xi t  ax t
2
• Gives final position in terms of velocity and acceleration
• Doesn’t tell you about final velocity
KINEMATIC EQUATIONS, 5
• For constant a,
v xf2  v xi2  2ax  xf  xi 
• Gives final velocity in terms of acceleration and
displacement
• Does not give any information about the time
KINEMATIC EQUATIONS SUMMARY
TABLE 2-1
• 𝑣𝑥 = 𝑣𝑥𝑖 + 𝑎𝑥 𝑡
• 𝑥 = 𝑥𝑖 + 𝑣𝑥𝑖 𝑡 +
2
• 𝑣𝑥2 = 𝑣𝑥𝑖
+ 2𝑎𝑥 ∆𝑥
• ∆𝑥 =
𝑣𝑥𝑖 +𝑣𝑥
2
𝑡
1
𝑎𝑥 𝑡 2
2
GRAPHICAL LOOK AT MOTION WITH
CONSTANT ACCELERATION: X-T CURVE
• The slope of the curve is the
velocity
xf  xi  v xi t 
1 2
ax t
2
• The curved line indicates
the velocity is changing
– Therefore, there is an
acceleration
GRAPHICAL LOOK AT MOTION:
VELOCITY – TIME CURVE
• The slope gives the
acceleration
• 𝑣𝑥 = 𝑣𝑥𝑖 + 𝑎𝑥 𝑡
• The straight line indicates a
constant acceleration
GRAPHICAL LOOK AT MOTION:
ACCELERATION – TIME CURVE
• The acceleration is constant.
FREELY FALLING OBJECTS
• A freely falling object is any object moving freely under
the influence of gravity alone
• The acceleration of an object in free fall is directed
downward, regardless of the initial motion
ACCELERATION OF FREELY FALLING
OBJECT
• The magnitude of free fall acceleration is g = 9.80 m/s2
– g decreases with increasing altitude
– g varies with latitude
– 9.80 m/s2 is the average at the Earth’s surface
– The italicised g will be used for the acceleration due to
gravity
• We will neglect air resistance
FREE FALL – AN OBJECT DROPPED
• Initial velocity is zero
• Let up be positive
vo= 0
• Use the kinematic equations
– Generally use y instead of x
since vertical
• Acceleration is
ay = -g = -9.80 m/s2
a = -g
FREE FALL – AN OBJECT THROWN
DOWNWARD
• ay = -g = -9.80 m/s2
• Initial velocity  0
– With upward being
positive, initial velocity will
be negative
vo= 0
a = -g
FREE FALL – OBJECT THROWN
UPWARD
• Initial velocity is upward, so
positive
• The instantaneous velocity at
the maximum height is zero
• ay = -g = -9.80 m/s2
everywhere in the motion
•The motion may be symmetrical
– Then tup = tdown , v = -vo
v=0
vo≠ 0
a = -g
FREE FALL – KINEMATIC EQUATIONS
• 𝑣𝑦 = 𝑣𝑦𝑖 − 𝑔𝑡
• 𝑦=
1
𝑦𝑖 + 𝑣𝑦𝑖 𝑡 − 𝑔𝑡 2
2
2
• 𝑣𝑦2 = 𝑣𝑦𝑖
− 2𝑔∆𝑦
• ∆𝑦 =
𝑣𝑦𝑖 +𝑣𝑦
2
𝑡
KINEMATIC EQUATIONS FROM
CALCULUS
• Displacement equals the area
under the velocity – time
curve
lim
tn 0
v
n
tf
xn
tn   v x (t )dt
ti
• The limit of the sum is a
definite integral
KINEMATIC EQUATIONS – GENERAL
CALCULUS FORM
dv x
ax 
dt
t
v xf  v xi   ax dt
0
dx
vx 
dt
t
xf  xi   v x dt
0
KİNEMATİC EQUATİONS – FROM
INTEGRATİON
• The integration form of vf – vi gives
v xf  v xi  ax t
• The integration form of xf – xi gives
1
xf  xi  v xi t  a x t 2
2
PROBLEM SOLVING – CONCEPTUALIZE
• Think about and understand the situation
• Make a quick drawing of the situation
• Gather the numerical information
– Include algebraic meanings of phrases
• Focus on the expected result.
– Think about units
• Think about what a reasonable answer should be
PROBLEM SOLVING – CATEGORISE
• Simplify the problem
– Can you ignore air resistance?
– Model objects as particles
• Classify the type of problem
– Substitution
– Analysis
• Try to identify similar problems you have already solved
– What analysis model would be useful?
PROBLEM SOLVING – ANALYZE
• Select the relevant equation(s) to apply
• Solve for the unknown variable
• Substitute appropriate numbers
• Calculate the results
– Include units
• Round the result to the appropriate number of significant
figures
PROBLEM SOLVING – FINALIZE
• Check your result
– Does it have the correct units?
– Does it agree with your conceptualised ideas?
• Look at limiting situations to be sure the results are
reasonable
• Compare the result with those of similar problems