January 23, Week 2physics.unm.edu/Courses/morgan-tracy/160/Slides/160-01-23-13.pdf · 1/23/2013...

Velocity January 23, 2013 - p. 1/12

January 23, Week 2

Today: Chapter 2, Instantaneous Velocity and Accleration

Homework Assignment #2 - Due January 25

Mastering Physics: 6 problems from chapters 1 and 2.

Written Question: 2.75

Box numbers will be posted online tomorrow morning

Friday office hours will be held in Regener Hall 114


Review

Position, x: - How far and what direction from origin

0

x


Review


0

x

Displacement, ∆x = x2 − x1 - Change in Position


Review


0

x


Distance, d - how far from initial to final position, alwayspositive


Review


0

x



Average Speed: spav =d

∆t


Review


0

x



Average Speed: spav =d

∆t

Average Velocity: vav =∆x

∆t=

x2 − x1

t2 − t1unit: m/s


Average Velocity Exercise

Consider three cars which perform the following motions:Car #1 travels to the right at 15m/s.

Car #2 goes to the right for 5 s at 10m/s, immediately turnsaround, and goes to the left at 25m/s for 2 s.

Car #3 travels 90m to the left every minute.

Which of the following is the correct ranking of the cars’average velocities from smallest to largest? (Use the usualsign convention and put negative numbers first.)







(a) #1, #2, #3







(a) #1, #2, #3 (b) #1, #3, #2







(a) #1, #2, #3 (b) #1, #3, #2

(c) #3, #1, #2







(a) #1, #2, #3 (b) #1, #3, #2

(c) #3, #1, #2 (d) #3, #2, #1







(a) #1, #2, #3 (b) #1, #3, #2

(c) #3, #1, #2 (d) #3, #2, #1

(e) #2, #1, #3


Velocity Exercise Followup









Car #1: vav = 15m/s






Car #1: vav = 15m/s

Car #2:It travels (10m/s)(5 s) = 50m to the right and then(25m/s)(2 s) = 50m to the left. Its displacement forthe entire trip is 0 ⇒ vav = 0






Car #1: vav = 15m/s

Car #2:It travels (10m/s)(5 s) = 50m to the right and then(25m/s)(2 s) = 50m to the left. Its displacement forthe entire trip is 0 ⇒ vav = 0

Car #3: vav =−90m

1min=

−90m

60 s= −1.5m/s


Significant Figures

■ Significant Figures = express the accuracy of ameasurement.


Significant Figures


■ Usually just the number of digits you see in the number.


Significant Figures


■ Usually just the number of digits you see in the number.■ Exceptions:

◆ Strings of zeros at the end of large numbers or at thebeginning of small numbers are not significant.

◆ Zeroes at the end of all numbers are significant.


Significant Figures




◆ Zeroes at the end of all numbers are significant.■ When multiplying or dividing, we round to the fewest number

of significant figures.


Significant Figures




◆ Zeroes at the end of all numbers are significant.■ When multiplying or dividing, we round to the fewest number

of significant figures.■ When adding or subtracting, we round to the fewest places

past the decimal point.


Significant Figures Exercise

A car travels from x = −70 km to x = −57 km in 7minutes.What is the car’s average velocity, in kilometers per minute,recorded to the proper number of significant figures?




(a)13 km

7min= 1.857 km/min




(a)13 km

7min= 1.857 km/min (b)

−13 km

7min= −1.857 km/min




(a)13 km


−13 km


(c)−70 km

7min= −10 km/min




(a)13 km


−13 km


(c)−70 km

7min= −10 km/min (d)

−57 km

7min= −8 km/min




(a)13 km


−13 km


(c)−70 km

7min= −10 km/min (d)

−57 km

7min= −8 km/min

(e)13 km

7min= 2 km/min


Unit Conversion

We use the fact that when multiplying or dividing physicalquantities that their units also multiply and divide to simplifyunit conversion.


Unit Conversion

We use the fact that when multiplying or dividing physicalquantities that their units also multiply and divide to simplifyunit conversion.Given that 1.00 in = 2.54 cm, which of the following is thecorrect conversion of 5.00 cm/s into in/hr?


Unit Conversion


(a)5 cm

s×

1 in

2.54 cm×

3600 s

1h= 7090 in/hr


Unit Conversion


(a)5 cm

s×

1 in

2.54 cm×

3600 s

1h= 7090 in/hr

(b)5 cm

s×

2.54 cm

1 in×

3600 s

1h= 45700 in/hr


Unit Conversion


(a)5 cm

s×

1 in

2.54 cm×

3600 s

1h= 7090 in/hr

(b)5 cm

s×

2.54 cm

1 in×

3600 s

1h= 45700 in/hr

(c)5 cm

s×

1 in

2.54 cm×

60 s

1h= 118 in/hr


Unit Conversion


(a)5 cm

s×

1 in

2.54 cm×

3600 s

1h= 7090 in/hr

(b)5 cm

s×

2.54 cm

1 in×

3600 s

1h= 45700 in/hr

(c)5 cm

s×

1 in

2.54 cm×

60 s

1h= 118 in/hr

(d)5 cm

s×

1 in

2.54 cm×

1h

3600 s= 5.47× 10−4 in/hr


Unit Conversion


(a)5 cm

s×

1 in

2.54 cm×

3600 s

1h= 7090 in/hr

(b)5 cm

s×

2.54 cm

1 in×

3600 s

1h= 45700 in/hr

(c)5 cm

s×

1 in

2.54 cm×

60 s

1h= 118 in/hr

(d)5 cm

s×

1 in

2.54 cm×

1h

3600 s= 5.47× 10−4 in/hr

(e) None of the these


Velocity

Average velocity: vav =∆x

∆t


Velocity


∆t

Average velocity is a good starting point, but it’s not sufficientfor most physics problems, since it only tells you what happenson average.


Velocity


∆t

Average velocity is a good starting point, but it’s not sufficientfor most physics problems, since it only tells you what happenson average.

Instantaneous velocity vx - how fast and the direction of motionfor one instant of time.


Motion Graphs

Physicists like to make graphs to describe motion.


Motion Graphs


Position versus time


Motion Graphs


Position versus time Velocity versus time


Motion Graphs


xPosition versus time Velocity versus time


Motion Graphs


t

xPosition versus time Velocity versus time


Motion Graphs


t

xPosition versus time

vxVelocity versus time


Motion Graphs


t


t



Motion Graphs


t


t


Horizontal Motion


Motion Graphs


t


t


Horizontal Motion

Vertical Motion


Motion Graphs


t


t


Horizontal Motion

Vertical Motion

t

yPosition versus time


Motion Graphs


t


t


Horizontal Motion

Vertical Motion

t

yPosition versus time

t

vyVelocity versus time


Uniform Motion Position Graph

Uniform Motion - Constant velocity

b b b b




Walking to right motion diagram:

b b b b




Walking to right motion diagram: b b b b





Equal spacing between dots because with constant velocity theobject travels the same distance during equal elapsed times.






t

x Position versus time






t

x Position versus time In uniform motion,position is a straightline






t


t1 t2






t


t1 t2

x1






t


t1 t2

x1

x2






t


t1 t2

x1

x2

t2 − t1 = ∆t






t


t1 t2

x1

x2

t2 − t1 = ∆t

x2 − x1 = ∆x






t


t1 t2

x1

x2

t2 − t1 = ∆t

x2 − x1 = ∆x

vx =∆x

∆t






t


t1 t2

x1

x2

t2 − t1 = ∆t

x2 − x1 = ∆x

vx =∆x

∆t

Velocity is the slope of the position versus time graph


Math and Physics Slopes

In Physics, slopes have units and don’t necessarily correspondto the steepness of the line on the drawing.




In math, the slope of line tellsyou how "steep" a line is.




0 1 2 30

1

2





0 1 2 30

1

2

run

rise





0 1 2 30

1

2

run

rise


Slope: m =rise

run=

1

1= 1




0 1 2 30

1

2

run

rise


Slope: m =rise

run=

1

1= 1

In Physics, the slope of line isthe ratio of the change in twophysical quantities.




0 1 2 30

1

2

run

rise


Slope: m =rise

run=

1

1= 1

t(s)

x(m)





0 1 2 30

1

2

run

rise


Slope: m =rise

run=

1

1= 1

t(s)

x(m)

1 2 3

15

30





0 1 2 30

1

2

run

rise


Slope: m =rise

run=

1

1= 1

t(s)

x(m)

1 2 3

15

30

∆t = 1 s

∆x = 15m





0 1 2 30

1

2

run

rise


Slope: m =rise

run=

1

1= 1

t(s)

x(m)

1 2 3

15

30

∆t = 1 s

∆x = 15m


Slope = Velocity: vx =∆x

∆t=

15m

1 s= 15m/s


Position Graph Exercise

A man walks some distance to the right with constant speed,immediately turns around and walks back to his starting pointwith the same speed. Which of the following is the correctposition-versus-time graph?

Motion Diagram:b b b b

b b b b Same Point





b b b b Same Point

t

x(a)





b b b b Same Point

t

x(a)

t

x(b)





b b b b Same Point

t

x(a)

t

x(b)

t

x(c)





b b b b Same Point

t

x(a)

t

x(b)

t

x(c)

t

x(d)





b b b b Same Point

t

x(a)

t

x(b)

t

x(c)

t

x(d)

t

x(e)

January 23, Week 2physics.unm.edu/Courses/morgan-tracy/160/Slides/160-01-23-13.pdf · 1/23/2013...

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Transcript of January 23, Week 2physics.unm.edu/Courses/morgan-tracy/160/Slides/160-01-23-13.pdf · 1/23/2013...