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Motion in a straight line Average, and instantaneous velocity Average, and instantaneous acceleration All found in University Physics, 2.2-2.4 Lecture 2 Today… weekendnotes.com
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Page 1: Newton's Laws of Motion L2.1

Motion in a straight line

• Average, and instantaneous velocity

• Average, and instantaneous acceleration

All found in University Physics, 2.2-2.4

Lecture 2

Today… weekendnotes.com

Page 2: Newton's Laws of Motion L2.1
Page 3: Newton's Laws of Motion L2.1

ReviewLast lecture we looked at :• The difference between a vector and a scalar quantity.

Vector notation.

• The distinction between distance and displacement. Distance is a scalar quantity. Displacement is a vector

• Addition and subtraction of vectors

• Components of vectors

• Unit vectors. The scalar (dot) product

Page 4: Newton's Laws of Motion L2.1

Example- Adding vectors• Find A + B using the component method if A has

magnitude 5.0m at elevation of 600 and B - magnitude 3.0m at elevation of 300 as shown below.

mji

jmm

imm

jBAiBABAR yyxx

ˆ8.5ˆ1.5

ˆ)30sin0.360sin0.5(

ˆ)30cos0.360cos0.5(

ˆ)(ˆ)(

00

00

A

B

300

300

5m

3m

5.8m

5.1m

X

Y

Page 5: Newton's Laws of Motion L2.1

011

22

7.481.5

8.5tantan

7.7)8.5()1.5(magnitudeResultant

m

m

R

RAngle

mmm

x

y

5.8m

5.1m

Example- Adding vectors (cont)

5.8m

5.1m

X X

Y Y

θ

Page 6: Newton's Laws of Motion L2.1

Example- Calculation of components

5.8m

5.1m

X

Y

θ=120°

A

0

0

120sin

120cos

AA

AA

y

x

θ1=30° 0

0

30cos

30sin

AA

AA

y

x

θ1=30°

000000

000000

30sin90cos30cos90sin3090sin

30sin90sin30cos90cos3090cos

AAA

AAA

y

x

Page 7: Newton's Laws of Motion L2.1

The Displacement vectorand

Distance scalar

If P2 is at P1 then displacement is zero but distance travelled still has a value

Average velocity =

displacement/time

Average speed =

distance/time

Page 8: Newton's Laws of Motion L2.1

Speed and velocity

Speed is measured using distance and time:

Velocity is measured using displacement and time:

t

x

tt

xxvav

12

12

t

d

tt

dd

12

12speed

So speed is a scalar, velocity is a vector.For example 15 km/hour is a scalar. 58.4 m/s in a northerly direction is a velocity and is a vector.

Page 9: Newton's Laws of Motion L2.1

Example - Average velocity vs speed Ian Thorpe’s personal best for 100 m freestyle in a 50 m pool in 2001 was 48.81 seconds.

His average SPEED would have been 2.05 m/s

His average VELOCITY however is zero m/s!webswimming.tripod.com

Page 10: Newton's Laws of Motion L2.1

Average velocity

t

x

tt

xxav

12

12v

Average velocity is the displacement divided by the time interval

Page 11: Newton's Laws of Motion L2.1

t

x

tt

xxav

12

12vx2=1680 mt2 = 85 sec

sms

m

t

xav /23

73

1675v

Example- Find the average velocity.

x1=5 mt1= 12 sec

stop

Only includes initial and final conditions, the average does not have details about the path or events. Such as stopping or drive over the mountain.

old.risk.ru

Page 12: Newton's Laws of Motion L2.1

Average velocity from graph

Page 13: Newton's Laws of Motion L2.1

Example - Average velocity

It takes Penny 37.0 seconds to drive along a 1.2 km airport runway, from a standing start at one end to the other end. What is her average velocity in completing this displacement.

To become a velocity this speed needs to include a direction.Average velocity v=32.4m/s in a due North direction.

hunternissan.com

Page 14: Newton's Laws of Motion L2.1

Instantaneous velocity

t

x

tt

xxav

12

12vvelocityAverage

Instantaneous velocity is the limit of the average velocity as the time interval approaches zero. It is the instantaneous rate of change of position with time

dt

dx

t

x

t

lim

0instantv

howstuffworks.com

Page 15: Newton's Laws of Motion L2.1

Example 2.1 Average and instantaneous velocity

mtx 2520

ischeetah ofPosition

Page 16: Newton's Laws of Motion L2.1

Example 2.1Average and Instantaneous velocity (cont)

a) Find the displacement of the cheetah during the interval between t = 1.0 sec and t = 2.0 sec.

vav=Δx/Δt=(40 m– 25m) /(2s – 1s)=15 m/s in the positive x direction

b) Find the average velocity during this time interval.

mtx 2520

Page 17: Newton's Laws of Motion L2.1

Example 2.1Average and Instantaneous velocity (cont)

sms

m

t

xSo

mxst

mxst

st

av /5.101.0

05.1v

05.26;1.1

;25;1

1.0

22

11

c) Find the instantaneous velocity at time t = 1.0 sec, by taking Δt = 0.1 sec, then Δt = 0.01 sec.

d) Derive a general expression for the instantaneous velocity as a function of time, and from it find v, at t = 1.0 sec and t = 2.0 sec.

sms

m

t

xSo

mxt

mxst

statst

av /05.1001.0

1005.0v

1005.25;s 01.1

25 ;1

1 01.0 Using

22

11

1 nn nttdt

d

mtx 2520

Page 18: Newton's Laws of Motion L2.1

Instantaneous velocity from a graph

As P2 approaches P1 the slope of the line between P1 and P2 becomes the tangent at point P1 and its slope gives the value for instantaneous velocity at P1 ….. On a graph of position as a function of time for straight-line motion, the instantaneous velocity at any point is equal to the slope of the tangent to the curve at that point

Page 19: Newton's Laws of Motion L2.1

Average acceleration

t

x

tt

xxav

12

12v

Average velocity is the displacement divided by the time interval

Average acceleration is the change in velocity divided by the time interval

tttaav

vvv

12

12

Units of acceleration are m/s2

Page 20: Newton's Laws of Motion L2.1

Instantaneous acceleration

Instantaneous acceleration is the limit of the average acceleration as the time interval approaches zero; it is the instantaneous rate of change of velocity with time

tttaav

vvv

onaccelarati Average12

12

2

2

0instant

vv

dt

xd

dt

dx

dt

d

dt

d

ta

t

lim

Page 21: Newton's Laws of Motion L2.1

Example2.3- Instantaneous acceleration

smt /50.060v

by given iscar theof velocity The2

Page 22: Newton's Laws of Motion L2.1

Example 2.3 Instantaneous acceleration (cont)

a) Find the change in velocity of the car in the time interval between t =1.0 sec and t = 3.0 sec.

At t = 1 sec..v = 60.5 m/s At t = 3sec..v = 64.5 m/s

b) Find the average acceleration in this time interval

smt /50.060v 2

Page 23: Newton's Laws of Motion L2.1

Example 2.3 Instantaneous acceleration (cont)

c) Find the average acceleration at time t=1.0 s by taking Δt to be first 0.1 s, then 0.01 s.

2/05.11.0

5.60605.60v

1.0 using 1timeAt

sms

s

m

s

m

ta

stst

av

d) Derive an expression for the instantaneous acceleration at any time, and use it to find the acceleration at t = 1.0 s and t = 3.0 s.

2/005.101.0

5.6051005.60v

01.0using 1timeAt

sms

s

m

s

m

ta

stst

av

1 nn nttdt

d

smt /50.060v 2

Page 24: Newton's Laws of Motion L2.1

Instantaneous acceleration from a graph

Page 25: Newton's Laws of Motion L2.1

Review: What does the displacement vs time graph look like?

Page 26: Newton's Laws of Motion L2.1

Review: What does the velocity vs time graph look like?

Page 27: Newton's Laws of Motion L2.1

Example: A dizzy turkey• A turkey moves according to the following

relationship: x = 5 – 2 t + 0.5 t2 (m)• Calculate its instantaneous velocity and

acceleration at a time t = 5 s.

Differentiate: v = -2 + t (m/s); a = 1 m/s2

At t = 5 sv = 3 m/s; acceleration always 1m/s2

simpleist.com

Page 28: Newton's Laws of Motion L2.1

Lets look at an object falling from afar.

An astronaut floats near the space shuttle. His acceleration is a function of his position.

wikipedia.org

Page 29: Newton's Laws of Motion L2.1

legacy.shadowlordinc.com

However around the surface of the Earth the acceleration is fairly constant so a velocity time graph is a straight line.

vm/s

t sec

Page 30: Newton's Laws of Motion L2.1

Remember…

• Average velocity=Δx/ Δt

• Instantaneous Velocity = dx/dt as Δt0….

… the derivative of x w.r.t t

• Average acceleration = Δv/ Δt

• Instantaneous acceleration = dv/dt as Δt0…. …the derivative of v w.r.t t

starwars.com

Page 31: Newton's Laws of Motion L2.1

Lecture 2-What you need to know and where to find it

• Average, and instantaneous velocity

• Average, and instantaneous acceleration

All found in University Physics, 2.2-2.4