1. Linear Kinematics
Objectives:
• Define the idea of spatial reference frames
• Introduce the concepts of position,
displacement, distance, velocity, and speed
• Learn how to compute displacement,
velocity, and speed
• Learn the difference between average,
instantaneous, and relative velocity
Linear Kinematics
Kinematics
• The form, pattern, or sequencing of movement
with respect to time
• Forces causing the motion are not considered
Linear Motion
• All parts of an object or system move the same
distance in the same direction at the same time
Linear Kinematics
• The kinematics of particles, objects, or systems
undergoing linear motion
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2. Spatial Reference Frames
• A spatial reference frame is a set of coordinate axes
(1, 2, or 3), oriented perpendicular to each other.
• It provides a means of describing and quantifying
positions and directions in space
1-dimensional (1-D) Reference Frame
• Quantifies positions and directions along a line
unit of measure
origin
label
x = 2m
x (m)
-5 -4 -3 -2 -1 0 1 2 3 4 5
– direction + direction
2-D Reference Frame
• Quantifies positions and directions in a plane
y (m)
(x,y) = (1 m, 2 m)
2
+y direction
–y direction
unit of measure
label
90° θ=63°
(0,0) x (m)
-2 2
origin +x direction
–x direction
2

3. 3-D Reference Frame
• Quantifies positions and directions in space
z (m)
2
(x,y,z) = (1.5m, 2m, 1.8m)
(0,0,0)
φ=36°
y (m)
2
θ=53°
x (m) 2
Reference Frames & Motion
• The position or motion of a system does not
depend on the choice of reference frame
• But, the numbers used to describe them do
• Always specify the reference frame used!
y1 (m)
)
(m
x2
)
(m
y2
1
1
1
x1 (m)
1
3

4. Selecting a Reference Frame
• Use only as many dimensions as necessary
• Align reference frame with fixed, clearly-defined,
and physically meaningful directions.
(e.g. compass directions, anatomical planes)
Common 2-D conventions:
– Sagittal plane: +X = anterior; +Y = upward
– Transverse plane: +X = left; +Y = upward
– Horizontal plane: +X = anterior; +Y = left
• The origin should also have physical meaning.
2-D examples:
– X=0 initial position
– Y=0 ground height
Position
• The location of a point, with respect to the origin,
within a spatial reference frame
• Position is a vector; has magnitude and direction
• Or, specify position by the coordinates of the point
• Position has units of length (e.g. meters, feet)
y (m) (x,y) = (1m, 2m)
2
Point’s position is:
• distance of 2.24m at an angle
2.24m of 63° above the +x axis, or
• (x,y) position of (1m, 2 m)
origin
θ=63°
x (m)
(0,0) 2
4

5. Linear Displacement
• Change (directed distance) from a point’s initial
position to its final position
• Displacement is a vector; has magnitude and direction
• Displacement has units of length (e.g. meters, feet)
y (m)
initial position
displacement
1 pinitial final position
pfinal
x (m)
1
Computing Displacement
• Compute displacement (∆p) by vector subtraction
∆p = pfinal – pinitial
y (m) initial position
1
pinitial final position
pfinal
x (m)
1
–pinitial –pinitial
∆p
5

6. Describing Displacement
• Can describe displacement by:
– Magnitude and direction
(e.g. 2.23m at 26.6° below the +x axis)
– Components y (m)
(change) along
each axis 1 pinitial
(e.g. 2m in the +x
pfinal
direction, 1m in
1
the –y direction)
θ ∆px x (m)
∆py
∆p
-1
Distance
• The length of the path traveled between a point’s
initial and final position
• Distance is a scalar; it has magnitude only
• Has units of length (e.g. meters, feet)
• Distance ≥ (Magnitude of displacement)
Displacement
= 64 m West
E
N
Distance = 200m
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7. Example Problem #1
A box is resting on a table of height 0.3m.
A worker lifts the box straight upward to a
height of 1m.
He carries the box straight backward 0.5m,
keeping it at a constant height.
He then lowers the box straight downward to
the ground.
What was the displacement of the box?
What distance was the box moved?
Linear Velocity
• The rate of change of position
• Velocity is a vector; has magnitude and direction
change in position displacement
velocity = =
change in time change in time
• Shorthand notation:
pfinal – pinitial ∆p
v = =
tfinal – tinitial ∆t
• Velocity has units of length/time (e.g. m/s, ft/s)
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8. Computing Velocity
• direction of velocity = direction of displacement
• magnitude of velocity = magnitude of displacement
change in time
• component of velocity = component of displacement
change in time
y ∆p y V = ∆p / ∆t
t
/∆
∆p
∆p
∆py vy = ∆py / ∆t
=
θ v θ
x x
∆px vx = ∆px / ∆t
Speed
• The distance traveled divided by the time taken
to cover it
• Equal to the average magnitude of the
instantaneous velocity over that time.
distance
speed =
change in time
• Speed is a scalar; has magnitude only
• Speed has units of length/time (e.g. m/s, ft/s)
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9. Speed vs. Velocity
Displacement
= 64 m West
E
N
Distance = 200m
Assume a runner takes 25 s to run 200 m:
200 m 64 m West
Speed = Velocity =
25 s 25 s
= 8 m/s = 2.6 m/s West
Example Problem #2
During the lifting task of example problem #1, it
takes the worker 0.5s to lift the box, 1.3s to
carry it backward, and 0.6s to lower it.
What were the average velocity and average
speed of the box during the first, lifting phase
of the task?
What were the average velocity and average
speed of the box for the task as a whole?
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10. Velocity as a Slope
• Graph x-component of position vs. time
• x-component of velocity from t1 to t2
= slope of the line from px at t1 to px at t2
Slope : ∆px / ∆t = vx
px (m)
∆px
∆t
t1 t2 time (s)
Average vs. Instantaneous Velocity
• The previous formulas give us the average velocity
between an initial time (t1) and a final time (t2)
• Instantaneous velocity is the velocity at a single
instant in time
• Can estimate instantaneous velocity using the
central difference method:
p (at t1 + ∆t) – p (at t1 – ∆t)
v (at t1) =
2 ∆t
where ∆t is a very small change in time
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11. Instantaneous Velocity as a Slope
• Graph of x-component of position vs. time
slope = instantaneous
x-velocity at t1
px (m)
slope = average
x-velocity from
t1 to t2
∆t
t1 t2 time (s)
Estimating Velocity from Position
Identify points with
px (m)
zero slope = points
with zero velocity
Portions of the curve
with positive slope
time (s) have positive velocity
(i.e. velocity in the
vx (m/s)
+ direction)
Portions of the curve
with negative slope
0
time (s) have negative velocity
(i.e. velocity in the
– direction)
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12. Relative Position
• Find the position of one point or object relative to
another by vector subtraction of their positions
p(2 relative to 1) = p2 – p1
p2 = p1 + p(2 relative to 1)
y (m)
object 2
1 p(2 relative to 1)
p2 object 1
p1
x (m)
1
Relative Velocity
• Apparent velocity of a second point or object to
an observer at a first moving point or object
• Compute by vector subtraction of the velocities
v(2 relative to 1) = v2 – v1
vy (m/s)
v2 = v1 + v(2 relative to 1)
object 2
1 v(2 relative to 1)
v2 object 1
v1
vx (m/s)
1
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13. Example Problem #3
A runner on a treadmill is running at 3.4 m/s in
a direction 10° left of forward, relative to the
treadmill belt (resulting in a forward velocity
of 3.3 m/s and a leftward velocity of 0.6 m/s,
relative to the belt).
The treadmill belt is moving backward at 3.6
m/s.
What is the runner’s overall velocity?
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