11 Lecture Ppt Rectilinear Motion

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Tenth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Phillip J. Cornwell

Lecture Notes:

Brian P. SelfCalifornia Polytechnic State University

CHAPTER

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

11Kinematics of Particles-

Rectilinear motion


Vector Mechanics for Engineers: Dynamics

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Contents

11 - 2

Introduction

Rectilinear Motion: Position,

Velocity & Acceleration

Determination of the Motion of a

Particle

Sample Problem 11.2

Sample Problem 11.3

Uniform Rectilinear-Motion

Uniformly Accelerated Rectilinear-

Motion

Motion of Several Particles:

Relative Motion

Sample Problem 11.4

Motion of Several Particles:

Dependent Motion

Sample Problem 11.5

Graphical Solution of Rectilinear-

Motion Problems

Curvilinear Motion: Position, Velocity

& Acceleration

Derivatives of Vector Functions

Rectangular Components of Velocity

and Acceleration

Motion Relative to a Frame in

Translation



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Application

11 - 3

Kinematic relationships are used to

help us determine the trajectory of a

golf ball, the orbital speed of a

satellite, and the accelerations

during acrobatic flying.



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Introduction

11 - 4

• Dynamics includes:

Kinetics: study of the relations existing between the forces acting on

a body, the mass of the body, and the motion of the body. Kinetics is

used to predict the motion caused by given forces or to determine the

forces required to produce a given motion.

Kinematics: study of the geometry of motion.

Relates displacement, velocity, acceleration, and time without reference

to the cause of motion.Fthrust

Flift

Fdrag



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Introduction

11 - 5

• Particle kinetics includes:

• Rectilinear motion: position, velocity, and acceleration of a

particle as it moves along a straight line.

• Curvilinear motion: position, velocity, and acceleration of a

particle as it moves along a curved line in two or three

dimensions.



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Rectilinear Motion: Position, Velocity & Acceleration

11 - 6

• Rectilinear motion: particle moving

along a straight line

• Position coordinate: defined by

positive or negative distance from a

fixed origin on the line.

• The motion of a particle is known if

the position coordinate for particle is

known for every value of time t.

• May be expressed in the form of a

function, e.g., 326 ttx

or in the form of a graph x vs. t.



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11 - 7

• Instantaneous velocity may be positive or

negative. Magnitude of velocity is referred

to as particle speed.

• Consider particle which occupies position P

at time t and P’ at t+Dt,

t

xv

t

x

t D

D

D

D

D 0lim

Average velocity

Instantaneous velocity

• From the definition of a derivative,

dt

dx

t

xv

t

D

D

D 0lim

e.g.,

2

32

312

6

ttdt

dxv

ttx



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11 - 8

• Consider particle with velocity v at time t and

v’ at t+Dt,

Instantaneous accelerationt

va

t D

D

D 0lim

tdt

dva

ttv

dt

xd

dt

dv

t

va

t

612

312e.g.

lim

2

2

2

0

D

D

D

• From the definition of a derivative,

• Instantaneous acceleration may be:

- positive: increasing positive velocity

or decreasing negative velocity

- negative: decreasing positive velocity

or increasing negative velocity.



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Concept Quiz

11 - 9

What is true about the kinematics of a particle?

a) The velocity of a particle is always positive

b) The velocity of a particle is equal to the slope of

the position-time graph

c) If the position of a particle is zero, then the

velocity must zero

d) If the velocity of a particle is zero, then its

acceleration must be zero



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11 - 10

• From our example,

326 ttx

2312 ttdt

dxv

tdt

xd

dt

dva 612

2

2

- at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0

- at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2

• What are x, v, and a at t = 2 s ?

• Note that vmax occurs when a=0, and that the

slope of the velocity curve is zero at this point.

• What are x, v, and a at t = 4 s ?



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Determination of the Motion of a Particle

11 - 11

• We often determine accelerations from the forces applied

(kinetics will be covered later)

• Generally have three classes of motion

- acceleration given as a function of time, a = f(t)

- acceleration given as a function of position, a = f(x)

- acceleration given as a function of velocity, a = f(v)

• Can you think of a physical example of when force is a

function of position? When force is a function of velocity?

a spring drag



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Acceleration as a function of time, position, or velocity

11 - 12

a a t 0 0

v t

v

dv a t dt ( )dv

a tdt

v dv a x dx

0 0

v x

v x

v dv a x dx a a x

and dx dv

dt av dt

dv

v a vdx

0 0

v t

v

dvdt

a v

0 0

x v

x v

v dvdx

a v

a a v

( )dv

a vdt

If…. Kinematic relationship Integrate



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Sample Problem 11.2

11 - 13

Determine:

• velocity and elevation above ground at

time t,

• highest elevation reached by ball and

corresponding time, and

• time when ball will hit the ground and

corresponding velocity.

Ball tossed with 10 m/s vertical velocity

from window 20 m above ground.

SOLUTION:

• Integrate twice to find v(t) and y(t).

• Solve for t when velocity equals zero

(time for maximum elevation) and

evaluate corresponding altitude.

• Solve for t when altitude equals zero

(time for ground impact) and evaluate




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Sample Problem 11.2

11 - 14

tvtvdtdv

adt

dv

ttv

v

81.981.9

sm81.9

00

2

0

ttv

2s

m81.9

s

m10

2

21

00

81.91081.910

81.910

0

ttytydttdy

tvdt

dy

tty

y

2

2s

m905.4

s

m10m20 ttty

SOLUTION:

• Integrate twice to find v(t) and y(t).



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Sample Problem 11.2

11 - 15

• Solve for t when velocity equals zero and evaluate

corresponding altitude.

0s

m81.9

s

m10

2

ttv

s019.1t

• Solve for t when altitude equals zero and evaluate


22

2

2

s019.1s

m905.4s019.1

s

m10m20

s

m905.4

s

m10m20

y

ttty

m1.25y



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Sample Problem 11.2

11 - 16

• Solve for t when altitude equals zero and evaluate


0s

m905.4

s

m10m20 2

2

ttty

s28.3

smeaningles s243.1

t

t

s28.3s

m81.9

s

m10s28.3

s

m81.9

s

m10

2

2

v

ttv

s

m2.22v



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Sample Problem 11.3

11 - 17

Brake mechanism used to reduce gun

recoil consists of piston attached to barrel

moving in fixed cylinder filled with oil.

As barrel recoils with initial velocity v0,

piston moves and oil is forced through

orifices in piston, causing piston and

cylinder to decelerate at rate proportional

to their velocity.

Determine v(t), x(t), and v(x).

kva

SOLUTION:

• Integrate a = dv/dt = -kv to find v(t).

• Integrate v(t) = dx/dt to find x(t).

• Integrate a = v dv/dx = -kv to find

v(x).



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Sample Problem 11.3

11 - 18

SOLUTION:

• Integrate a = dv/dt = -kv to find v(t).

000

ln

v t

v

v tdv dva kv k dt kt

dt v v

ktevtv 0

• Integrate v(t) = dx/dt to find x(t).

0

0 0

00 0

1

kt

tx t

kt kt

dxv t v e

dt

dx v e dt x t v ek

ktek

vtx 10



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Sample Problem 11.3

11 - 19

• Integrate a = v dv/dx = -kv to find v(x).

kxvv

dxkdvdxkdvkvdx

dvva

xv

v

0

00

kxvv 0

• Alternatively,

0

0 1v

tv

k

vtx

kxvv 0

00 or

v

tveevtv ktkt

ktek

vtx 10with

and

then



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Problem Solving

11 - 20

A bowling ball is dropped from a boat so that it

strikes the surface of a lake with a speed of 5 m/s.

Assuming the ball experiences a downward

acceleration of a =10 - 0.01v2 when in the water,

determine the velocity of the ball when it strikes the

bottom of the lake.

Which integral should you choose?

0 0

v t

v

dv a t dt 0 0

v x

v x

v dv a x dx

0 0

v t

v

dvdt

a v

0 0

x v

x v

v dvdx

a v

(a)

(b)

(c)

(d)

+y



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Problem Solving

11 - 21

SOLUTION:

• Determine the proper kinematic

relationship to apply (is acceleration

a function of time, velocity, or

position?

• Determine the total distance the car

travels in one-half lap

• Integrate to determine the velocity

after one-half lap

The car starts from rest and accelerates

according to the relationship

23 0.001a v

It travels around a circular track that has

a radius of 200 meters. Calculate the

velocity of the car after it has travelled

halfway around the track. What is the

car’s maximum possible speed?



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Problem Solving

11 - 22

dv

v a vdx

0 0

x v

x v

v dvdx

a v

Choose the proper kinematic relationship

Given:23 0.001a v

vo = 0, r = 200 m

Find: v after ½ lap

Maximum speed

Acceleration is a function of velocity, and

we also can determine distance. Time is not

involved in the problem, so we choose:

Determine total distance travelled

3.14(200) 628.32 mx r



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Problem Solving

11 - 23

0 0

x v

x v

v dvdx

a v

Determine the full integral, including limits

628.32

2

0 03 0.001

vv

dx dvv

Evaluate the interval and solve for v

2

0

1628.32 ln 3 0.001

0.002

v

v

2628.32( 0.002) ln 3 0.001 ln 3 0.001(0)v

2ln 3 0.001 1.2566 1.0986= 0.15802v

2 0.158023 0.001v e Take the exponential of each side



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Problem Solving

11 - 24

2 0.158023 0.001v e Solve for v

0.158022 3

2146.20.001

ev

46.3268 m/sv

How do you determine the maximum speed the car can reach?

23 0.001a v Velocity is a maximum when

acceleration is zero

This occurs when20.001 3v

30.001maxv

max 54.772 m/sv



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Uniform Rectilinear Motion

11 - 25

For a particle in uniform

rectilinear motion, the

acceleration is zero and

the velocity is constant.

vtxx

vtxx

dtvdx

vdt

dx

tx

x

0

0

00

constant

During free-fall, a parachutist

reaches terminal velocity when

her weight equals the drag

force. If motion is in a straight

line, this is uniform rectilinear

motion.

Careful – these only apply to

uniform rectilinear motion!



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Uniformly Accelerated Rectilinear Motion

11 - 26

If forces applied to a body

are constant (and in a

constant direction), then

you have uniformly

accelerated rectilinear

motion.

Another example is free-

fall when drag is negligible



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Uniformly Accelerated Rectilinear Motion

11 - 27

For a particle in uniformly accelerated rectilinear motion, the

acceleration of the particle is constant. You may recognize these

constant acceleration equations from your physics courses.

0

0

0

constant

v t

v

dva dv a dt v v at

dt

0

210 0 0 0 2

0

x t

x

dxv at dx v at dt x x v t at

dt

0 0

2 2

0 0constant 2

v x

v x

dvv a v dv a dx v v a x x

dx

Careful – these only apply to uniformly

accelerated rectilinear motion!



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Motion of Several Particles

11 - 28

We may be interested in the motion of several different particles,

whose motion may be independent or linked together.



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Motion of Several Particles: Relative Motion

11 - 29

• For particles moving along the same line, time

should be recorded from the same starting

instant and displacements should be measured

from the same origin in the same direction.

ABAB xxx relative position of B

with respect to AABAB xxx

ABAB vvv relative velocity of B

with respect to AABAB vvv

ABAB aaa relative acceleration of B

with respect to AABAB aaa



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Sample Problem 11.4

11 - 30

Ball thrown vertically from 12 m level

in elevator shaft with initial velocity of

18 m/s. At same instant, open-platform

elevator passes 5 m level moving

upward at 2 m/s.

Determine (a) when and where ball hits

elevator and (b) relative velocity of ball

and elevator at contact.

SOLUTION:

• Substitute initial position and velocity

and constant acceleration of ball into

general equations for uniformly

accelerated rectilinear motion.

• Substitute initial position and constant

velocity of elevator into equation for

uniform rectilinear motion.

• Write equation for relative position of

ball with respect to elevator and solve

for zero relative position, i.e., impact.

• Substitute impact time into equation

for position of elevator and relative

velocity of ball with respect to

elevator.



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Sample Problem 11.4

11 - 31

SOLUTION:

• Substitute initial position and velocity and constant

acceleration of ball into general equations for

uniformly accelerated rectilinear motion.

2

2

221

00

20

s

m905.4

s

m18m12

s

m81.9

s

m18

ttattvyy

tatvv

B

B

• Substitute initial position and constant velocity of

elevator into equation for uniform rectilinear motion.

ttvyy

v

EE

E

s

m2m5

s

m2

0



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Sample Problem 11.4

11 - 32

• Write equation for relative position of ball with respect to

elevator and solve for zero relative position, i.e., impact.

025905.41812 2 ttty EB

s65.3

smeaningles s39.0

t

t

• Substitute impact time into equations for position of elevator

and relative velocity of ball with respect to elevator.

65.325Ey

m3.12Ey

65.381.916

281.918

tv EB

s

m81.19EBv



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Motion of Several Particles: Dependent Motion

11 - 33

• Position of a particle may depend on position of one

or more other particles.

• Position of block B depends on position of block A.

Since rope is of constant length, it follows that sum of

lengths of segments must be constant.

BA xx 2 constant (one degree of freedom)

• Positions of three blocks are dependent.

CBA xxx 22 constant (two degrees of freedom)

• For linearly related positions, similar relations hold

between velocities and accelerations.

022or022

022or022

CBACBA

CBACBA

aaadt

dv

dt

dv

dt

dv

vvvdt

dx

dt

dx

dt

dx



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Sample Problem 11.5

11 - 34

Pulley D is attached to a collar which

is pulled down at 75 mm/s. At t = 0,

collar A starts moving down from K

with constant acceleration and zero

initial velocity. Knowing that velocity

of collar A is 300 mm/s as it passes L,

determine the change in elevation,

velocity, and acceleration of block B

when block A is at L.

SOLUTION:

• Define origin at upper horizontal surface

with positive displacement downward.

• Collar A has uniformly accelerated

rectilinear motion. Solve for acceleration

and time t to reach L.

• Pulley D has uniform rectilinear motion.

Calculate change of position at time t.

• Block B motion is dependent on motions

of collar A and pulley D. Write motion

relationship and solve for change of block

B position at time t.

• Differentiate motion relation twice to

develop equations for velocity and

acceleration of block B.



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Sample Problem 11.5

11 - 35

SOLUTION:

• Define origin at upper horizontal surface with

positive displacement downward.

• Collar A has uniformly accelerated rectilinear

motion. Solve for acceleration and time t to reach L.

( )0

2

mm mm300 225 1.333 s

s s

= +

= =

A A Av v a t

t t

22

0 0

2

2

2

mm mm300 2 200mm 225

s s

A A A A A

A A

v v a x x

a a



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Sample Problem 11.5

11 - 36

• Pulley D has uniform rectilinear motion. Calculate

change of position at time t.

xD = xD( )0+ vDt

xD - xD( )0= 75

mm

s

æ

èç

ö

ø÷ 1.333s( ) =100 mm

• Block B motion is dependent on motions of collar

A and pulley D. Write motion relationship and

solve for change of block B position at time t.

Total length of cable remains constant,

0 0 0

0 0 0

0

2 2

2 0

200mm 2 100mm 0

A D B A D B

A A D D B B

B B

x x x x x x

x x x x x x

x x

( )0

400mmB Bx x- = -



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Sample Problem 11.5

11 - 37

• Differentiate motion relation twice to develop

equations for velocity and acceleration of block B.

2 constant

2 0

mm mm300 2 75 0

s s

A D B

A D B

B

x x x

v v v

v

mm mm450 450

s sBv = - =

2

2 0

mm225 2(0) 0

s

A D B

B

a a a

a

2 2

mm mm225 225

s sBa = - =



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Problem Solving

11 - 38

Slider block A moves to the left with a

constant velocity of 6 m/s. Determine the

velocity of block B.

Solution steps

• Sketch your system and choose

coordinate system

• Write out constraint equation

• Differentiate the constraint equation to

get velocity



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Problem Solving

11 - 39

Given: vA= 6 m/s left Find: vB

xA

yB

This length is constant no

matter how the blocks move

Sketch your system and choose coordinates

Differentiate the constraint equation to

get velocity

const nts3 aA Bx y L

Define your constraint equation(s)

6 m/s + 3 0Bv

2 m/sB v

Note that as xA gets bigger, yB gets smaller.



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Graphical Solution of Rectilinear-Motion Problems

11 - 40

Engineers often collect position, velocity, and acceleration

data. Graphical solutions are often useful in analyzing

these data.Data Fideltity / Highest Recorded Punch

0

20

40

60

80

100

120

140

160

180

47.76 47.77 47.78 47.79 47.8 47.81

Time (s)

Accele

rati

on

(g

)

Acceleration data

from a head impact

during a round of

boxing.



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11 - 41

• Given the x-t curve, the v-t curve is

equal to the x-t curve slope.

• Given the v-t curve, the a-t curve is

equal to the v-t curve slope.



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11 - 42

• Given the a-t curve, the change in velocity between t1 and t2 is

equal to the area under the a-t curve between t1 and t2.

• Given the v-t curve, the change in position between t1 and t2 is

equal to the area under the v-t curve between t1 and t2.

11 Lecture Ppt Rectilinear Motion

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