MMB222_KinematicsParticles-01-OneDimension lecture 1

7/28/2019 MMB222_KinematicsParticles-01-OneDimension lecture 1

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MMB222 Dynamics

Kinematics of particle:

Position, displacement, velocity, acceleration

One dimensional motion Graphical methods

Prof. Jacek Uziak


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Kinematics deals with the concepts needed todescribemotion

Displacement or position: Dimension [L], SIunit meter(m)

Velocity: Dimension [L/T], SI unitmeters/second (m/s)

Acceleration: Dimension [L/T2

], SI unitmeters/second-squared (m/s2)

(Dynamics deals with the effect that forces

have on motion)


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First consideration: one dimensional motion(position, velocity, acceleration)

Note: in one dimension, there is little practical differencebetween vectors and scalars, so often people dispense withvector notation. However, to help remember what the conceptualnature of vector objects are, full vector notation will be used asmuch as possible.

To be able to find body position being given

its acceleration & starting position orDetermine acceleration if position is given as

a function of time


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x

PO

x

The motion of a particlealong a straight line istermed rectilinear motion.To define the position Pofthe particle on that line,we choose a fixed origin Oand a positive direction.

The distance xfrom Oto P, with theappropriate sign, completely defines theposition of the particle on the line and is called

theposition coordinateof the particle.

One Dimensional Motion


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Displacement

Displacement is a vector that pointsfrom an objects initial to final position,and whose magnitude is the distancebetween initial and final positions.

xix

fx

if xxxorigin


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Notation

We typically (but not always!) precede asymbol with a delta (

) to denote that it isthe change, or differencein the quantity weare referring to

examples:if xxx

12ttt

ba vvv


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Average Speed and Velocity

i.e. the average speed is the magnitudeof the average velocity vector

average speed =total distance

elapsed time

average velocity =displacement vector

elapsed time

12

12

tt

xx

t

xv

12

12

ttxx

txv


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Example

The space shuttle travels at an averagespeed of 7.6x103 m/s. The blink of aneye is roughly 100 ms. How far does theshuttle travel in the blink on an eye?

Answerm760


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Instantaneous velocity

the instantaneous velocity of an objectis its velocity vector at a given instantin time

can be calculated by measuring theaverage displacement over a small timeinterval t, and taking the limit as t

goes to zero:

t

xv

t 0lim

Limits are the starting point

in the mathematics of

calculus. Calculus is needed

to handle continuously

changing velocities.


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PO

xx

The instantaneous velocity vof the particleis equal to the time derivative of the positioncoordinate x,

v =dx

dt


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v =dx

dtPO

xx

+-

The velocity vis represented by algebraicnumber which can be positive or negative.

A positive value for vindicates that theparticle moves in the positive direction,and a negative value that it moves in thenegative direction.


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Example: graphical analysis of velocityA bus makes a trip

according to the followingposition vs. time diagram.

a) What is the finaldisplacement of the bus?Answer:

b) what is theinstantaneous velocity at3 hours after the start ofthe trip?

Answer:

c) what is the averagevelocity over the first 3hours of the trip?

Answer:

kmdf 10

)(kmx

0 1 2 3 4 5 6

60

50

40

30

20

10

0

)(hourst

hkmvht

/103

!0


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Acceleration

acceleration is the change of velocitywith respect to timeacceleration is a vector: it has a magnitude

and a directionaverage acceleration:

instantaneous acceleration:

12

12

tt

vv

t

vaavg

t

va

tinst

0lim


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PO

xx

The instantaneous acceleration ais obtainedby differentiating vwith respect to t,

a =dv

dtor a =

d2x

dt2

We can also express aas

a = vdv

dx


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PO

xx

+-

The acceleration ais represented by

algebraic number - positive or negative.A positive value for a, means that the

particle is truly accelerated (i.e., moves

faster) in the positive direction.A negative value for ais subject to a similar

interpretation.

a =dv

dta =

d2x

dt2

a = vdv

dx


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Acceleration

Note that the acceleration and velocity of anobject can have differentmagnitudes anddirections at the same instant of timea stopped car at a red light has zero acceleration

and zero velocitywhen the light goes green and the gas pedal is

pressed, the instant before starting to move thecar has zero velocity but a positive acceleration

driving at a constant velocity, the car has zeroacceleration

slowing down to stop at a red light, the car has anegative acceleration (i.e. deceleration) butpositive velocity


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Equations of kinematics for constantacceleration

Given an object moving with a constantacceleration, we want to know itskinematical properties

acceleration(easy, its a constant!)velocityposition

as a function of time


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Two types of motion are frequentlyencountered:uniform rectilinear motion, in which thevelocity vof the particle is constant and

x

=x

o +vt

and uniformly accelerated rectilinear motion,in which the acceleration a of the particle isconstant and

v = vo + at

v2 = vo

+ 2a(x -xo

)2

x =xo + vot + at21

2


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Example

A speeding motorist zips past a cop at200km/h. The cop, initially moving at50km/h, gives chase and accelerates

with a constant acceleration of 1m/s2.i) how long does it take the cop to catch

the motorist? Answer:

ii) how fast is the cop going when he/shedoes catch up? Answer:

83.34s

350 km/h (!)


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Example

A spacecraft is traveling at a constantvelocity of 3250m/s. It suddenly fires itsretrorockets, and starts decelerating at

10m/s2

. What is the velocity of the craftwhen its displacement is 215km relative tothe place where it first fired theretrorockets?

Answer: sm /105.2 3Note: 2 answers! Does this

make sense!!? (yes)


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Gravity and freely falling objects

The earth exerts a gravitational force on objects

the magnitude of this force is inverselyproportional tothe squareof the distance from the center of the

earth.

however, we are quite far from the center of the earth(6400km), so the difference in the magnitude of theforce between, say sea level, and Mount Everest (9km

above sea level), is negligiblefor most purposes

therefore by Newtons 2nd law: force = mass xacceleration, since the force of gravity is effectively aconstant, so is the acceleration due to this force


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Gravity and freely falling objects

Therefore, the kinematics of an object in free-fallnear the surface of the earth is governed by theequations of constant acceleration

The local magnitude of the acceleration due togravity is traditionally denoted by g, and is:

The direction of the acceleration is towards thecenter of the earth

2/81.9 smg Note: 3 significantfigures!


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Example

A rock is dropped from the top of a 10story building (30m).

i) How long does it take to reach theground? Answer:

ii) how fast is it going when it hits theground? Answer:

2.47s

24.23 m/s


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ExampleStanding next to the edge of a cliff, a person

fires two shots from a gun, one (a) straightupward, the other, (b) straight downward. Thegun shoots pellets at 30m/s from the barrel.

Ignoring air resistance, in which scenario (a) or(b) does the pellet hit the ground at the bottom

of the cliff with the greatest velocity?


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Graphical analysis of velocity andacceleration Consider graphs of the position or velocity of an

object as a function of time: x=x(t), v=v(t)

Notice that in these constant velocity/accelerationcases, the velocity is the slope of the x vs. tgraph, and the acceleration is the slope of the v vs.t graph.

Constant velocity example Constant acceleration example


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Graphical analysis of velocity andacceleration

In more general cases, this observationremains true:

instantaneousvelocity at time t is theslope, or tangent line of the trajectory x(t)(position vs. time) at the time t

instantaneousacceleration at time t is theslope, or tangent line of the curve v(t)(velocity vs. time) at the time t


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The graphical solution involves x- t, v- t,and a- tcurves.

a

tt1 t2

v2 - v1 = a dtt1

t2

x

tt1 t2

x1

x2

v

t

v1

v2

t1 t2

x2 -x1 = v dtt1

t2

At any given time t,v= slope of x- tcurvea= slope of v- tcurve

while over any given timeinterval t1 to t2,

v2 - v1 = area under a- tcurvex2 - x1 = area under v- tcurve

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MMB222_KinematicsParticles-01-OneDimension lecture 1