PHYS 1443 – Section 003 Lecture #4 - UTA HEP

Monday, Feb. 1, 2021 PHYS 1443-003, Spring 2021 Dr. Jaehoon Yu

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PHYS 1443 – Section 003Lecture #4

Monday, Feb. 1, 2021Dr. Jaehoon Yu

• Concepts of kinematic quantities• Average Velocity and Speed• Instantaneous velocity and speed• Acceleration

Today’s homework is homework #3, due 11pm, Tuesday, Feb. 9!!


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Announcements• First term exam in class Wednesday, Feb. 10

– BYOF: You may bring a one 8.5x11.5 sheet (front and back) of handwrittenformulae and values of constants for the exam

– No derivations, word definitions, setups or solutions of any problems, figures, pictures, diagrams or arrows, etc!

– No additional formulae or values of constants will be provided!– Must email me the photos of front and back of the formula sheet, including

the blank at [email protected] no later than 12:00pm the day of the test• The subject of the email should be the same as your file name• File name must be FS-E1-LastName-FirstName-SP21.pdf• Once submitted, you cannot change, unless I ask you to delete part of the sheet!

• Quiz 1 Results– Class average: 45.1/70

• Equivalent to 64/100– Top score: 70/70

mailto:[email protected]

Monday, Feb. 1, 2021 3

Special Project #2 for Extra Credit• Show that the trajectory of a projectile motion is a parabola!! (20 points)

– You MUST show full details of your OWN computations, including every step of the derivation, to obtain any credit

• Beyond what was covered in this lecture note and in the book!• You must show your OWN work in detail to obtain the full credit

• Must be handwritten and in much more detail than in this lecture note or the book!• Please do not copy from the lecture note or from your friends. You will all get 0!!• BE SURE to show all the details of your own work, including all formulae, proper

references to them and explanations

• Due at the beginning of the class 1:00pm Monday, Feb. 15 on Canvas– File name must be: SP2-LastName-FirstName-SP21.pdf

PHYS 1443-003, Spring 2021 Dr. Jaehoon Yu


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Recap: Displacement, Velocity and SpeedOne dimensional displacement is defined as (poll6):

ixxx f -ºD

The average velocity is defined as: vx−avg ≡

The average speed is defined as: vavg ≡

Total Distance TraveledTotal Elapsed Time

f

f

i

i

x xt t-

=-

xt

DD

Displacement per unit time in the interval throughout the motion

Displacement is the difference between initial and final poisitions of the motion and is a vector quantity. How is this different than distance?

DisplacementElapsed Time

ºUnit (poll3, 6)?? m/s

Unit (poll3)? m

Unit (poll3, 6)? m/s

A vector quantity

A scalar quantity

A vector quantity

Can someone tell me what the difference between speed and velocity is?


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How much is the elapsed time? t=D t 0- t

1x 2x

What is the displacement? =xD 2x 1- x

What is the average velocity?

What is the average speed ? vx−avg =

x2 − x1

t − t0

=

vavg =

x2 − x1

t − t0

=

ΔxΔt

ΔxΔt

Monday, Feb. 1, 2021 66

Difference between Speed and Velocity• Let’s take a simple one dimensional translational

motion that has many segments:Let’s call this line as X-axis

Let’s have 6 segments of motions in a total time interval of 20 s.

+10m +15m

-15m-5m -10m

+5m

Total Displacement: xD

Total Distance Traveled: D =

Average Velocity: f

f

ix

i

x xvt t-

º-

Average Speed: vavg ≡

Total Distance TraveledTotal Time Interval

xt

D=D

020

= 0( / )m s=

≡ xf − xi i ix x= - 0( )m=

10 15 5 15 10 5 60( )m+ + + + + =6020

= 3( / )m s=


Monday, Feb. 1, 2021 77

Example

2 1f ix x x x xD º - = -• Displacement:

• Average Velocity: f

f

ix

i

x xvt t-

º-

• Average Speed: Total Distance Traveled

Total Time Intervalv º

The position of a runner as a function of time is plotted as moving along the x axis of a coordinate system. During a 3.00-s time interval, the runner’s position changes from x1=50.0m to x2=30.5 m, as shown in the figure. What was the runner’s average velocity? What was the average speed?

30.5 50.0= - 19.5( )m= -

2 1

2 1

x x xt t t- D

= =- D

19.5 6.50( / )3.00

m s-= = -

50.0 30.5 19.5 6.50( / )3.00 3.00

m s- += = = +


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Example of a segmented motionYou drove a beat-up pickup truck along a straight road for 8.4km at 70km/h, at which point the truck ran out of gas and stopped. Over the next 30min, you walked another 2.0km farther along the road in the same direction to a gas station.


(a) what is your overall displacement from the beginning of your drive to your arrival at the station? You can split this motion in two segments

Segment 2: from the point where you started walking to the gas station, Δx2

Segment 1: from the start to the point where the truck ran out of gas, Δx1 Δx1 ≡ xf − xi =

Δx2 ≡ xf − xi =Overall displacement Δx is Δx =

+8.4km

+2.0km

Δx1 + Δx2 = +8.4+ +2.0( ) =

10.4 km( ) = 10400 m( )

Monday, Feb. 1, 2021 99

Example, cnt’dYou drove a beat-up pickup truck along a straight road for 8.4km at 70km/h, at which point the truck ran out of gas and stopped. Over the next 30min, you walked another 2.0km farther along the road in the same direction to a gas station.


(b) What is the time interval Δt from the beginning of your drive to your arrival at the station? Figure out the time interval in each segments of the motion and add them together. We know what the interval for the second segment is: Δt2=30minWhat is the interval for the first segment, Δt1? Use the definition of average velocity and solve it for Δt1

vx1 =

Δx1

Δt1Add the two time intervals for total time interval Δt

⇒Δt1 = 8.470

= 0.12 hr( ) =

7.2 min( )

Δx1

vx1

=

Δt = Δt1 + Δt2 = 7.2+ 30 = 37 min( )

= 37 × 60 = 2200 s( )

Monday, Feb. 1, 2021 1010

Example, cnt’dYou drove a beat-up pickup truck along a straight road for 8.4km at 70km/h, at which point the truck ran out of gas and stopped. Over the next 30min, you walked another 2.0km farther along the road in the same direction to a gas station.


Using the definition of the average velocity, just simply divide the overall displacement by the overall time interval:

(c) What is your average velocity vavg from the beginning of your drive to your arrival at the station?

vx = ΔxΔt

=Δx1 + Δx2Δt1 + Δt2

=+10.4 km( )37 min( ) =

+10400 m( )2200 s( ) = +4.7 m s( )

What is the average speed?What is the average velocity and average speed in km/h?


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Instantaneous Velocity and Speed• Can average quantities tell you the detailed story of the

whole motion?

vx =

Δt→0lim

ΔxΔt

= dxdt

*Magnitude of Vectors are Expressed in absolute values

•Instantaneous speed is the size (magnitude) of the velocity vector (poll 6,3): :

vx =

Δt→0lim

ΔxΔt

= dxdt

• Instantaneous velocity is defined as (poll 6,3):– What does this mean?

• Displacement in an infinitesimal time interval• Average velocity over a very, very short amount of time

NO!!


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Instantaneous Velocity

Time

Average Velocity

Instantaneous Velocity


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Position vs Time Plot

timet1 t2 t3t=0

Position

x=0

x11 2 3

1. Running at a constant velocity (go from x=0 to x=x1 in t1, Displacement is + x1 in t1 time interval, velocity is)

2. Velocity is 0 (go from x1 to x1 no matter how much time changes)3. Running at a constant velocity but in the reverse direction as 1. (go

from x1 to x=0 in t3-t2 time interval, Displacement is - x1 in t3-t2 time interval)

It is helpful to understand motions to draw them on a position vs time plot.

Does this motion physically make sense?

v1=(x1-0)/(t1-0)=x1/t1


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151515

Example

2x =1x =

xv =

xD =

(a) Determine the displacement of the engine during the interval from t1=3.00s to t2=5.00s.

Displacement is, therefore:

A jet engine moves along a track. Its position as a function of time is given by the equation x=At2+B where A=2.10m/s2 and B=2.80m.

(b) Determine the average velocity during this time interval.

1 3.00tx = = 22.10 (3.00) 2.80 21.7m´ + =2 5.00tx = = 22.10 (5.00) 2.80 55.3m´ + =

2 1x x- = 55.3 21.7- = ( )33.6 m+

xt

D=

D33.6

5.00 3.00=

-( )33.6 16.8 /

2.00m s=


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Example cont’d

( ) 1-= nn nCtCtdtd

( )5.00xv t s= =

xv = At2=

Calculus formula for derivative

The derivative of the engine’s equation of motion is

(c) Determine the instantaneous velocity at t=t2=5.00s.

and ( ) 0=Cdtd

The instantaneous velocity at t=5.00s is

0limt

xtD ®

D=

Ddxdt

= ( )2d At Bdt

+

2 5.00A´ = ( )2.10 10.0 21.0 /m s´ =


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Definitions of Displacement, Velocity and Speed

dtdx

txvx =DD

=®lim

0Δt

Displacement (poll 6,5): ixxx f -ºD

Average velocity (poll 6,5):tx

ttxxvi

ix

f

f

DD

=--

º

Average speed (poll 6,5):Spent Time TotalTraveled Distance Total

ºv

Instantaneous velocity (poll 6,5):

Instantaneous speed(poll 6,5): dt

dxtxvx =DD

=®lim

0Δt


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Acceleration

• In calculus terms: The slope (derivative) of the velocity vector with respect to time or the change of slopes of position as a function of time

xa º xv ºanalogous to

xa ºdtdx

txvx =DD

º®lim

0Δtanalogous to

Change of velocity in time (what kind of quantity is this? Polls 6,5 )

•Definition of the average acceleration:

•Definition of the instantaneous acceleration:

xf

f

xi

i

v vt t-

=-

xvt

DD

f

f

i

i

x xt t-

=-

xt

DD

xvt®

D=

DΔt 0lim

xdvdt

= d dxdt dtæ ö =ç ÷è ø

2

2d xdt

Vector


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Acceleration vs Time Plot

What does this plot tell you?Yes, you are right!!The acceleration of this motion is a constant!!The speed increases at a constant rate!


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Example

xa = )/(2.40.521

0.5021 2sm==

-=

A car accelerates along a straight road from rest to 75km/h in 5.0s.

What is the magnitude of its average acceleration?

xfv =

xiv =

( ) )/(105.4100036002.4 24

2

hkm´=´

=

0 /m s

750003600

ms=21 /m s

xf xi

f i

v vt t-

=-

xvt

DD


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A few Important Things on Acceleration• When an object is moving in a constant velocity (v=v0), there is

no acceleration (a=0)– Is there any acceleration when an object is not moving?

• When an object is moving faster as time goes on, (v=v(t) ), acceleration is positive (a>0).– Incorrect, since the object might be moving in negative direction initially

• When an object is moving slower as time goes on, (v=v(t) ), acceleration is negative (a<0)– Incorrect, since the object might be moving in negative direction initially

• In all cases, velocity is positive, unless the direction of the movement changes.– Incorrect, since the object might be moving in negative direction initially

• Is there acceleration if an object moves in a constant speed but changes direction?

The answer is YES!!

Monday, Feb. 1, 2021 222222

Displacement, Velocity, Speed & Acceleration

dtdx

txvx =DD

=®lim

0Δt

Displacement ixxx f -ºD

Average velocity tx

ttxxvi

ix

f

f

DD

=--

º

Average speedSpent Time TotalTraveled Distance Total

ºv

Instantaneous velocity

Instantaneous speed dtdx

txvx =DD

=®lim

0Δt

xa º xvt®

D=

DΔt 0lim

xdvdt

= d dxdt dtæ ö =ç ÷è ø

2

2d xdtInstantaneous acceleration

Average acceleration xa ºxf

f

xi

i

v vt t-

=-

xvt

DD



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Example for Acceleration• Velocity, vx, is express in: • Find the average acceleration in time interval, t=0 to t=2.0s

vx (t)

•Find instantaneous acceleration at any time t and t=2.0s

vxi(ti = 0)

ax t( ) ax(t = 2.0)Instantaneous

Acceleration at any time

Instantaneous Acceleration at any time t=2.0s

= 40−5t2( )m / s

vxf (t f = 2.0)

ax

= 40(m / s)

= 40−5× 2.02( ) = 20(m / s)

=

vxf − vxi

t f − ti =Δvx

Δt = 20− 40

2.0− 0 = −10(m / s2 )

≡ dvx

dt = d

dt40−5t2( )

= −10t m s2( ) = −10× (2.0)

= −20(m / s2 )


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Example for Acceleration• Position is express in: • Find the particle’s velocity function v(t) and the

acceleration function a(t). x = 4− 27t + t3 m( )

•Find the average acceleration between t=2.0s and t=4.0s

vx (t) = dx

dt= d

dt4− 27t + t3( ) =

vx (t = 2) = −27 + 3t2 =

ax (t) = d 2x

dt2 =

vx (t = 4) = −27 + 3t2 = −27 + 3⋅ 4( )2

= +21 m s( )

−27 + 3t2

m s( )

+6t m s2( )

dvx

dt= d

dt−27 + 3t2( ) =

•Find the average velocity between t=2.0s and t=4.0s ax =

vx (t = 4)− vx (t = 2)4− 2

=

−27 + 3⋅ 2( )2

= −15 m s( )

21− −15( )2

= 362

= +18

m s2( )

PHYS 1443 – Section 003 Lecture #4 - UTA HEP

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Transcript of PHYS 1443 – Section 003 Lecture #4 - UTA HEP