wednesday, sept. 15, 2010 phys 1441-002, fall 2010 dr. jaehoon yu 1 phys 1441 – section 002...

Wednesday, Sept. 15, 2010

PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

1

PHYS 1441 – Section 002Lecture #4

Wednesday, Sept. 15, 2010Dr. Jaehoon Yu

• One Dimensional Motion• Instantaneous Velocity and Speed• Acceleration• Motion under constant acceleration• One dimensional Kinematic Equations• How do we solve kinematic problems?• Falling motions

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2PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu



4

Reminder: Special Problems for Extra Credit• Derive the quadratic equation for yx2-zx+v=0

5 points• Derive the kinematic equation

from first principles and the known kinematic equations 10 points

• You must show your OWN work in detail to obtain the full credit• Must be in much more detail than in pg. 19 of this lecture note!!!

• Due Monday, Sept. 27

2 2

0 02v v a x x



5

Displacement, Velocity and SpeedOne dimensional displacement is defined as:

ixxx f

The average velocity is defined as: xv

The average speed is defined as: Total Distance Traveled

Total Elapsed Timev

f

f

i

i

x x

t t

x

t

Displacement per unit time in the period throughout the motion

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

Displacement

Elapsed Time

Unit? m/s

Unit? m

Unit? m/s



6

Example 2.1

2 1f ix x x x x • Displacement:

• Average Velocity: f

f

ix

i

x xv

t t

• Average Speed:

Total Distance Traveled

Total Time Intervalv

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 x

t t t

19.5

6.50( / )3.00

m s

50.0 30.5 19.56.50( / )

3.00 3.00m s

Wednesday, Sept. 15, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu

7

How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m/s?

Average speed

Distance

Example Distance Run by a Jogger

Distance Elapsed time

Average speed Elapsed time

2.22 m s 5400 s 12000 m



8

Andy Green in the car ThrustSSC set a world record of 341.1 m/s in 1997. To establish such a record, the driver makes two runs through the course, one in each direction to nullify wind effects. From the data, determine the average velocity for each run.

Example: The World’s Fastest Jet-Engine Car

vr

t

xr

1609 m339.5m s

4.740 s

vr

t

xr

1609 m342.7 m s

4.695 s

What is the speed?

What is the speed?

v vr

v vr

339.5m s

342.7 m s

342.7 m s



9

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

the whole motion?

xx

vt

Δt 0lim

*Magnitude of Vectors are Expressed in absolute values

•Instantaneous speed is the size (magnitude) of the velocity vector:

xx

vt

Δt 0lim

• Instantaneous velocity is defined as:– What does this mean?

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

NO!!



10

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)

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 useful to understand motions to draw them on position vs time plots.



11



12

Velocity vs Time Plot



13

Displacement, Velocity and Speed

xx

vt

Δt 0lim

Displacement ixxx f

Average velocityt

x

tt

xxv

i

ix

f

f

Average speed Spent Time Total

Traveled Distance Totalv

Instantaneous velocity

Instantaneous speed xx

vt

Δt 0lim



14

Acceleration

xa xv analogs to

xa xx

vt

Δt 0limanalogs to

Change of velocity in time (what kind of quantity is this?)

•Average acceleration:

•Instantaneous acceleration: Average acceleration over a very short amount of time.

xf

f

xi

i

v v

t t

xv

t

f

f

i

i

x x

t t

x

t

xv

t

Δt 0

lim

Vector!

Unit? m/s2Dimension? [LT-2]



15

Acceleration vs Time Plot

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



16

Example 2.3

xa )/(2.40.5

21

0.5

021 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

)/(105.4

1000

36002.4 242

hkm

0 /m s

75000

3600

m

s21 /m s

xf xi

f i

v v

t t

xv

t



17

Few Confusing 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!!



18

One Dimensional Motion• Let’s focus on the simplest case: acceleration is a constant (a=a0)• Using the definitions of average acceleration and velocity, we can derive

equations of motion (description of motion, velocity and position as a function of time)

xa (If tf=t and ti=0) xfv

For constant acceleration, average velocity is a simple numeric average

xv

fx Resulting Equation of Motion becomes

xv

fx

xi xv a txa

xf

f

xi

i

v v

t t

xf xi

x

v va

t

2

xi xfv v

2

2

xi xv a t

1

2xi xv a t

f

f

i

i

x x

t t

f ixx x

vt

(If tf=t and ti=0) xix v t

xix v t 21

2i xi xx v t a t



19

One Dimensional Motion cont’d

fx

xa

Average velocity

t Since

fx

Substituting t in the above equation,

xv

2xfv

Solving for t

Resulting in

2

xi xfv vxix v t

2xi xf

i

v vx t

xf xiv v

t

xf xi

x

v x

a

2xf xi xf xi

ix

v v v vx

a

2 2

2xf xi

ix

v vx

a

2 2xi x f iv a x x



20

Kinematic Equations of Motion on a Straight Line Under Constant Acceleration

tavtv xxixf

2

2

1tatvxx xxiif

1 1

2 2f xi xf xix x v t v v t

ifxf xxavv xxi 222

Velocity as a function of time

Displacement as a function of velocities and time

Displacement as a function of time, velocity, and acceleration

Velocity as a function of Displacement and acceleration

You may use different forms of Kinetic equations, depending on the information given to you for specific physical problems!!

wednesday, sept. 15, 2010 phys 1441-002, fall 2010 dr. jaehoon yu 1 phys 1441 – section 002...

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