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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 consideredLinear Motion• All parts of an object or system move the same

distance in the same direction at the same timeLinear Kinematics• The kinematics of particles, objects, or systems

undergoing linear motion

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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

0-1 21 3 4 5-2-3-4-5

1-dimensional (1-D) Reference Frame• Quantifies positions and directions along a line

x (m)

unit of measureorigin

+ direction– direction

labelx = 2m

2-D Reference Frame• Quantifies positions and directions in a plane

+x direction–x direction

label

(0,0) x (m)

y (m)

2

2(x,y) = (1m, 2m)

unit of measure

origin

θ=63°

-2

+y d

irect

ion

–ydi

rect

ion

90°

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3-D Reference Frame• Quantifies positions and directions in space

(0,0,0)

x (m)

z (m)

2

2

(x,y,z) = (1.5m, 2m, 1.8m)

y (m)

2

θ=53°

φ=36°

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!

x1 (m)

y1 (m)

1

1

x 2(m

)

y 2(m

)

1

1

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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)

(0,0)x (m)

y (m)

2

2(x,y) = (1m, 2m)

originθ=63°

2.24m

Point’s position is:• distance of 2.24m at an angleof 63° above the +x axis, or

• (x,y) position of (1m, 2m)

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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)

x (m)

y (m)

1

1

initial position

final position

displacement

pinitial

pfinal

Computing Displacement• Compute displacement (∆p) by vector subtraction

x (m)

y (m)

1

1

initial position

final positionpinitialpfinal

∆p = pfinal – pinitial

–pinitial –pinitial∆p

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Describing Displacement

• Can describe displacement by:– Magnitude and direction

(e.g. 2.23m at 26.6° below the +x axis)– Components

(change) alongeach axis

x (m)

y (m)

1

1 pinitial

pfinal

∆p ∆py

∆pxθ

(e.g. 2m in the +x direction, 1m in the –y direction)

-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)

Distance = 200m

Displacement = 64 m West

N

E

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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

• Shorthand notation:

• Velocity has units of length/time (e.g. m/s, ft/s)

v =∆p∆t

velocity =change in position

change in timedisplacement

change in time=

pfinal – pinitial

tfinal – tinitial=

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Computing Velocity• direction of velocity = direction of displacement

• magnitude of velocity = magnitude of displacementchange in time

• component of velocity = component of displacementchange in time

V = ∆p / ∆t

xθ

y

v = ∆p

/ ∆t

vy = ∆py / ∆t

vx = ∆px / ∆t

∆p

xθ

y

∆p

∆px

∆py

Speed

• The distance traveled divided by the time taken to cover it

• Equal to the average magnitude of the instantaneous velocity over that time.

• Speed is a scalar; has magnitude only• Speed has units of length/time (e.g. m/s, ft/s)

speed =distance

change in time

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Speed vs. Velocity

Distance = 200m

Displacement = 64 m West

N

E

Assume a runner takes 25 s to run 200 m:

Speed = 200 m 25 s

= 8 m/s

Velocity = 64 m West25 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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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

time (s)

p x(m

)

∆t

t2t1

∆px

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:

v (at t1) = p (at t1 + ∆t) – p (at t1 – ∆t)

2 ∆t

where ∆t is a very small change in time

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Instantaneous Velocity as a Slope

• Graph of x-component of position vs. time

slope = average x-velocity from t1 to t2

time (s)

p x(m

)

t2t1

∆t

slope = instantaneous x-velocity at t1

Estimating Velocity from Position

time (s)

p x(m

)

time (s)

v x(m

/s)

0

Identify points with zero slope = points with zero velocity

Portions of the curve with positive slope have positive velocity (i.e. velocity in the + direction)

Portions of the curve with negative slope have negative velocity (i.e. velocity in the – direction)

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Relative Position• Find the position of one point or object relative to

another by vector subtraction of their positions

x (m)

y (m)

1

1

object 2

object 1p2p1

p(2 relative to 1) = p2 – p1

p(2 relative to 1)

p2 = p1 + p(2 relative to 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

vx (m/s)

vy (m/s)

1

1

object 2

object 1v2v1

v(2 relative to 1) = v2 – v1

v(2 relative to 1)

v2 = v1 + v(2 relative to 1)

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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?