preLECTURE 2 Ch2 F19 Velocity

DepartmentofPhysicsandAppliedPhysicsPHYS.1410Lecture2A.Danylov

Lecture 2

Chapter 1 & 2

Kinematics in One Dimension

Physics I

Course website:https://sites.uml.edu/andriy-danylov/teaching/physics-i/

I like speed!! Kinematics?

–Hmm!...Not so much


Distance/Displacement: Section 1.3

Speed/Average velocity: Section 1.4 Instantaneous velocity: Section 2.2

Average Acceleration: Section 1.5 Instantaneous Acceleration: Section 2.7

Today we are going to discuss:


Mechanics

There are three branches of Mechanics: Kinematics Motion Forces Statics Motion Forces Dynamics Motion Forces

Kinematics describes motion of objects we are not interested in reasons (forces) of a motion

Motion


KinematicsNeedtodevelopavocabularyfor

describingtranslationalmotion:

PositionLevel

VelocityLevel

AccelerationLevel


Frames of Reference

-5 5

-5

5 y-axis

x-axis

Beforestartingsolvinganyproblem,wehavetodefineacoordinatesystem(aframeofreference)todescribepositionandmotionofanobject

Inthisclass,wewillbaseproblemsinaCartesiancoordinatesystem.Wewillhave1Dand2Dproblems

Wehavefreedomtochoosedirectionofanaxis. For1dimensional(1D)motion(motioninastraightline),it’sbettertoalignthex‐axis

alongamotiondirection. Forfallingbodies,wetendtodescribepositionusingthey‐axis

origin

1 2 3 4 5

Positive direction+x

Negative direction-x

x

One dimensional (1D) problem Two dimensional (2D) problem

origin


Distance vs. Displacement


Distance vs. Displacement

Distance (scalar): the total path length traveled by an object

Displacement (vector): how far an object is from its starting point


Distance vs. Displacement (1D)Distanceisascalar

Displacementisavector– Avectorhasbothmagnitudeanddirection(orsignin1‐D)

Displacement =x2- x1=+40 m

Distance =

20 40 60 70

x2

X (m)

x1

70+30 =100 m

Displacement = final position – initial position


Distance vs. Displacement (1D)

x1 x2

Distance = 20 m

x1

Distance = 20 m

Displacement =negative

Displacement =

x1

30-10= + 20 m

x2

10-30= -20 m


Speed and Velocity

Even Hollywood feels that there is a difference between these two terms


Average Speed and Average Velocity

elapsed time travelleddistance

speedaverage

(Speed: Distance traveled per unit time interval)

Speed is a scalar

Velocity is a vector

(Velocity: Displacement of an object per unit time interval)

elapsed timentdisplaceme

velocityaverage


Graphs: Average velocity

0 1 2 3 4 5 60

5

10

15

20po

sitio

n (m

)

time (s)

∆t

∆x

t1 t2

x1

x2


Instantaneous Velocity

Average velocity does not tell the whole story…we also need:

If you watch a car’s speedometer, at any instant of time, the speedometer tells you how fast the car is going at that instant.


Instantaneous velocity

0 1 2 3 4 5 60

5

10

15

20po

sitio

n (m

)

time (s)

t1

goes to 0




Graphically, instantaneous velocity is the slope of the x vs t plot at a single point

Mathematically, the instantaneous velocity is the derivative of the position function


Silly example: Finding instantaneous velocity from position (graphically)

.

>0

Turning point

Moves forwardMoves backward

0

v positive

It flies back.v4 negative

v1 (slow speed)v2 (max speed)

v3=0 (turning point)

Steeper slope ≡ higher velocityGentler slope ≡ lower velocity

A test to find a turning point: velocity changes its sign


Finding Position from a Velocity Graph

The total displacement ∆s is called the “area under the velocity curve.” (the total area enclosed between the t-axis and the velocity curve).

The displacement is the shaded area

∆

Example

∙ Let’s integrate it: ∙ x ∙

∙

Initial position

Geometrical meaning of an integral is an area

Total displacement

displacement

v(t)

t


Acceleration


AccelerationVelocity can also change with time: acceleration

elapsed time velocityof change

onacceleratiaverage

Speeding up: acceleration

Instantaneous acceleration

If we are given x(t), we can find both velocity v(t) and acceleration a(t) as a function of time

Slowing down: deceleration


A particle is moving in a straight line so that its position is given by the relation

x = (2 m/s2)t2 + (3 m). Calculate

(a) its average acceleration during the timeinterval from t1 = 1 s to t2 = 2 s,

(b) its instantaneous acceleration as afunction of time.

Example 2-7: Acceleration given x(t).

Example 2-7 Acceleration given x(t)


Thank you

preLECTURE 2 Ch2 F19 Velocity

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