Chapter  2    

Mo,on  Along  a  Straight  Line  

One  dimensional  mo,on  

   

TYPICAL  SPEEDS  Motion v(mph) v(m/s) v/c Light 669,600,000 300,000,000 1 Earth around sun 66,600 29,600 10-4 Moon around Earth 2300 1000 3*10-6 Jet fighter 2200 980 3*10-6 Sound in air 750 334 10-6 Commercial airliner 600 267 10-6

Cheetah 62 28 10-7 Falcon diving 82 37 10-7 Olympic 100m dash 22 10 3*10-8 Flying bee 12 5 10-8 Walking ant 0.03 0.01 3*10-11 Swimming sperm 0.0001 0.000045 10-13

Nonrela,vis,c  speeds  

A  person  running  along  a  straight  line  at  some  velocity.  

x(m) -30 -20 -10 0 10 20 30

x1 =10 m x2=20 m

t1= 2 s t2= 4 s

Average velocity velocity = (distance traveled)/time v = (x2 ! x1) / (t2 ! t1)

v = !x / !t

x2 ! x1 = 20 !10 = 10mt2 ! t1 = 4 ! 2 = 2sv = 10m / 2s = 5m / s

Distance-­‐,me  graph  for  running  in  a  straight  line  Distance(m)

Time(s) 2 4 6

10

20

Average = Instantaneous Velocity or Exact

Δx

Δt

0

v = !x / !t

!x = v!tdistance = velocity " time

Constant  velocity-­‐,me  graph  

Velocity (m/s)

distance = velocity ! time = 5m / s ! 3s= 15m

area = width ! length

Time (s) 1 2 3

5

4

3

2

1

Area = distance = 15m !x = v!t

0  

Area = distance = 15m !x = v!tArea = distance = 15m !x = v!t

What  is  meant  by  vavg

Suppose I run for 5 s at a velocity of 2 m/s, then I rest for 5 s, and then I run for 10 s at a velocity of 2 m/s. What is my average velocity over the 20 s?

Distance-­‐,me  graph  for  changing  V  What  is  meant  by  vavg    

vavg = total displacement/total time

Total displacement = 10 + 0 + (30-10) = 30 m

Total time = 5 + (10 - 5) + (20 -10) = 20 s

= 30 / 20 = 1.50 m/s

Wrong Way

vav = (2m/s + 0m/s +2 m/s)/3

= 1.33 m/s

x(m)

t (secs) 5 10 15

10

20

30

20 0  

Distance-­‐,me  graph  for  changing  V  x(m)

t(secs) 5 10 15

10

20

30 vavg = total displacement/total time

= 30 / 20 = 1.50 m/s

20 0  

What  is  the  difference  between  average  velocity  and  average  speed?  

Suppose I run for 5 s at a velocity of 2 m/s, then I rest for 5 s, and then I run for 10 s at a velocity of 2 m/s. Now I run for 20 s at - 2m/s or backwards. What is my average velocity and speed over the entire 40 s?

Average velocity = vavg = total displacement/total time = (30 - 30) / 40 = 0 m/s

x(m)

t(secs) 5 10 15 20 25 30 35 40

10

20

30

Average speed = savg = (30 + 30)/40 = 1.5 m/s

Average velocity and average speed are not always the same.

0  

Here  is  the  velocity-­‐,me  graph  for  uniform  accelera,on.  

Units of a are (m/s2 in mks system of units)

t (secs)

2 v(m/s)

Δt

Δv

1

v = at

0  

a = velocity/time = (v2 ! v1) / (t2 ! t1) = "v / "t = average acceleration = slope of graph

How  far  does  an  object  move  from  point  1  to  point  2?  It  is  equal  to  the  total  area  under  the  green  line  in  between  points  1  and  2.  

Area =1/2 base x height + length x width

Area =12(t2 ! t1)(v2 ! v1) + v1(t2 ! t1) =

12(v2 + v1)(t2 ! t1)

x = vavg " t

v(m/s)

t (secs) 1 2 0  

NON-­‐ZERO  INITIAL  SPEED  

v

t

v0

v0 + at

v = v0 + at

x = vavt =12(v0 + v)t

x =12(v0 + v 0+at)t = v0t +

12at 2

0  

Summary  of  Equa,ons  in  1D  (constant  accelera,on)  

v = v0 + at

vavg =12(v0 + v)

x = vavt

x = v0t +12at 2

v2 = v02 + 2ax

Under  what  condi,ons  do  these  apply?    

Same  Equa,ons  with  ini,al  velocity  =  0  

v = at

vavg =12v

x = vavgt

x =12at 2

v2 = 2ax

Lets  look  at  a  numerical  example  and  then  a  demo.  

t =xvavg

t =2xa

Galileo’s  Result  (1564)  

Dropping  things  from  rest   Galileo’s experiments produced a surprising Result. All objects fall with the same acceleration Regardless of mass and shape. g = 9.8 m/s2 or 32 ft/s2

Neglecting air resistance.

Free  Fall  Example    

10 m

Find the time it takes for a free-fall drop from 10 m height.

Find

Take the downward direction as positive displacement. Use two methods.

t = 10 / vavgvavg

Method  1  Method  2  

t =2xa

x = 12 at

2x = vavgt

Free  Fall  Example    

10 m

Find the time it takes for a free-fall drop from 10 m height. Take the downward direction as positive displacement. Use two methods.

t = 10 / vavgvavg = (0 + vf ) / 2Find vfvf2 = 02 + 2gx

vf = 2 !9.8 !10vf = 14m / s

vavg =14m / s2

= 7m / s

t = 10m7m / s

t = 1.43s

vavg =12v f

Method  1  

+y ( Coordinate system in 1D)

x = 12 gt

2

t = 2y / g

t = 2 *10 / 9.8t =1.43 s

t = 2ya

Method  2  

Demos  (Mo,on  in  one  dimension)    

•  Find  the  ,me  between  hits  of  free  fall  accelera,on  of  three  weights  equally  spaced  on  a  string  50  cm  apart.  

•  The  ,mes  of  the  first  three  are:      

   

The  ,me  between  hits  on  the  floor  =    0.13  ,  0.10,  0.09    

y = 12 gt

2

t = 2y / g

t = 2 * 0.5 / 9.8t = 0.32 s

y = 12 gt

2

t = 2y / g

t = 2 *1.0 / 9.8t = 0.45 s

y = 12 gt

2

t = 2y / g

t = 2 *1.5 / 9.8t = 0.55 s

Time  between  hits  vrs  distance  

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 1 2 3 4 5

Distance from end of string

(meters)

Tim

e b

etw

een

pre

vio

us

hit

(secs)

Series1

How  do  you  space  the  weights  apart  such  that  they  hit  at  equal  successive  ,me  intervals?  

 

0

0.15

0.3

0.45

0.6

0.75

0.9

0 1 2 3

Distance from end of string

(meters)

Tim

e (s

ecs)

Series1

stone # Time Distance1 0.15 0.112 0.3 0.443 0.45 0.994 0.6 1.765 0.75 2.76

0

0.15

0.3

0.45

0.6

0.75

0.9

0 1 2 3

Distance from end of string

(meters)

Tim

e (s

ecs)

Series1

distance =12g(time)2

How  far  does  a  train  go  when  it  starts  from    rest  and  uniformly  increases  its  speed  to  120  m/s  in  1  

min?  

t = 0v = 0

v = at

x = vavgt

vavg =(0 +120m / s)

2= 60m / s

x = vavgt = (60m / s)! (60s) = 3600m

How  far  does  a  train  go  when  it  starts  from    rest  and  uniformly  increases  its  speed  to  v  m/s  in  ,me  t?  

v

t

t = 0v = 0

v = at

Find vavg

vavg =12(v0 + v) =

12(0 + at) = 1

2at

x = vavgt =12at ! t

x =12at 2

x =12(2m / s2 )(602 s) = 3600m

Find  the  ,me  t  it  takes  for  a  plaeorm  diver  10  m  high  to  hit  the  water  if  he  takes  off  ver,cally  with  a  velocity  of  -­‐  4  m/s  and  the  speed  v  with  which  the  diver  strikes  the  water.  

10 m

Non-zero Initial Velocity Example

v2 = v02 + 2ay

v = (!4)2 + (2)(9.8)(10)v =14.6m / s

vavg =(!4 +14.6)

2= 5.3m / s

t = 10mvavg

= 10m5.3m / s

=1.89s

Choose  posi,ve  y  down  

You and your dog go for a walk to the park. On the way, your dog takes many side trips to chase squirrels or examine fire hydrants. When you arrive at the park, do you and your dog have the same displacement?

1) yes

2) no

ConcepTest Walking the Dog

You and your dog go for a walk to the park. On the way, your dog takes many side trips to chase squirrels or examine fire hydrants. When you arrive at the park, do you and your dog have the same displacement?

1) yes

2) no

Yes, you have the same displacement. Since you and your dog had the same initial position and the same final position, then you have (by

definition) the same displacement.

ConcepTest Walking the Dog

Follow-up: Have you and your dog traveled the same distance?

You drive 4 miles at 30 mi/hr and then another 4 miles at 50 mi/hr. What is your average speed for the whole 8-mile trip?

1) more than 40 mi/hr

2) equal to 40 mi/hr

3) less than 40 mi/hr

ConcepTest Cruising Along

You drive 4 miles at 30 mi/hr and then another 4 miles at 50 mi/hr. What is your average speed for the whole 8-mile trip?

1) more than 40 mi/hr

2) equal to 40 mi/hr

3) less than 40 mi/hr

It is not 40 mi/hr! Remember that the average speed is distance/time.

Since it takes longer to cover 4 miles at the slower speed, you are actually moving at 30 mi/hr for a longer period of time! Therefore, your average speed is closer to 30 mi/hr than it is to 50 mi/hr.

ConcepTest Cruising Along

Follow-up: How much further would you have to drive at 50 mi/hr in order to get back your average speed of 40 mi/hr?

ConcepTest  Accelera,on  

 If  the  velocity  of  a  car  is  non-­‐

zero  (v  ≠0),  can  the  

accelera7on  of  the  car  be  

zero?  

1) Yes

2) No

3) Depends on the velocity

ConcepTest  2.8a  Accelera,on  I  

Sure it can! An object moving with constant velocity

has a non-zero velocity, but it has zero acceleration

since the velocity is not changing.

 If  the  velocity  of  a  car  is  

non-­‐zero  (v≠0),  can  the  

accelera7on  of  the  car  be  

zero?  

1) Yes

2) No

3) Depends on the velocity

 When  throwing  a  ball  

straight  up,  which  of  the  

following  is  true  about  its  

velocity  v  and  its  

accelera7on  a  at  the  highest  

point  in  its  path?  

1) both v = 0 and a = 0 2) v ¹ 0, but a = 0 3) v = 0, but a ≠0 4) both v ≠0 and a ≠ 0 5) not really sure

ConcepTest Acceleration

y At the top, clearly v = 0 because the ball has

momentarily stopped. But the velocity of the ball is changing, so its acceleration is definitely not zero! Otherwise it would remain at rest!!

 When  throwing  a  ball  straight  

up,  which  of  the  following  is  

true  about  its  velocity  v  and  its  

accelera7on  a  at  the  highest  

point  in  its  path?  

   

   

1) both v = 0 and a = 0 2) V ≠0, but a = 0 3) v = 0, but a ≠0 4) both v ≠ 0 and a ≠ 0 5) not really sure

ConcepTest Acceleration

Follow-up: …and the value of a is…?

¹ ¹

¹ ¹

A ball is thrown straight upward with some initial speed. When it reaches the top of its flight (at a height h), a second ball is thrown straight upward with the same initial speed. Where will the balls

cross paths?

1) at height h

2) above height h/2 3) at height h/2 4) below height h/2 but above 0 5) at height 0

ConcepTest 2.11 Two Balls in the Air

Ball 1:v2 tmax( ) = 0 = vinit2 + 2 !g( )h" h = vinit2 2g" vinit = 2gh

Define t = 0 when Ball 1at apex y1,0 = h( ) :y1 t( ) = h + v1,0t !

12gt 2 = h ! 1

2gt 2

y2 t( ) = 0 + v2,0t !12gt 2 = 2gh t ! 1

2gt 2

" at t ' balls at same height. i.e., y1 t '( ) = y2 t '( )

" h ! 12gt '2 = 2gh t '! 1

2gt '2 " t ' = h 2g

y1 t '( ) = y2 t '( ) = h ! 12g h 2g( )2 = h ! h4 = 3

4h

A ball is thrown straight upward with some initial speed. When it reaches the top of its flight (at a height h), a second ball is thrown straight upward with the same initial speed. Where will the balls

cross paths?

1) at height h 2) above height h/2 3) at height h/2 4) below height h/2 but above 0 5) at height 0

The first ball starts at the top with no initial speed. The second ball starts at the bottom with a large initial speed. Since the balls travel the same time until they meet, the second ball will cover more distance in that time, which will carry it over the halfway point before the first ball can reach it.

ConcepTest 2.11 Two Balls in the Air

Follow-up: How could you calculate where they meet?

Define the instantaneous velocity

Recall

(average) as Δt 0 = dx/dt (instantaneous) Example

Differential Calculus ���Definition of Velocity when it is smoothly changing

x = 12 at

2

x = f (t)

What is dxdt

?

v =(x2 ! x1)(t2 ! t1)

="x"t

v = lim !x!t

DISTANCE-TIME GRAPH FOR UNIFORM ACCELERATION

x

t

(t+Δt) t

v = Δx /Δt

x = f(t)

x + Δx = f(t + Δt)

dx/dt = lim Δx /Δt as Δt 0

. x, t

x = 12 at

2

x = f (t)

Differential Calculus: an example of a derivative

x = 12 at

2

x = f (t)

dx/dt = lim Δx /Δt as Δt 0

= f (t + !t) " f (t)!t

f (t) = 12 at

2

f (t + !t) = 12 a(t + !t)2

= 12 a(t

2 + 2t!t + (!t)2)

=12 a(t

2 + 2t!t + (!t)2) " 12 at

2

!t

=12 a(2t!t + (!t)2)

!t

= 12 a(2t + !t)

! at"t! 0

dxdt

= at velocity in the x direction

Integral Calculus How far does an object go under constant acceleration?

Distance equals area under speed graph regardless of its shape

Area = x = 1/2(base)(height) = 1/2(t)(at) = 1/2at2

v=dx/dt

t

v= at

x = vi!ti = ati!tii=1

N

"i=1

N

"

vi

Δti

tf

0

N =t f!ti

!x = v!t

Integration:anti-derivative

ati!tii=1

N

" = at0

t f# dt where !t i $ 0 and N $%

x = 12 at

2

at0

t f! dt = 12at 2

0

tf = 12a (tf

2 " 0) = 12a tf

2

y = cxn

dy / dx = ncxn!1Power Rule

Chain Rule

Product Rule y(x) = f (x)g(x)dydx

=dfdxg(x) + f (x) dg

dx

y(x) = y(g(x))dydx

=dydg

dgdx

y = 30x5

dydx

= 5(30)x4 = 150x4

y = 3x2 (ln x)dydx

= 2(3)x(ln x) + 3x2 (1x) = 6x ln x + 3x

dydx

= 3x(2 ln x +1)

y = (5x2 !1)3 = g3 where g=5x2 !1dydx

= 3g2 dgdx

= 3(5x2 !1)2 (10x)

dydx

= 30x(5x2 !1)2

Three Important Rules of Differentiation

Some Derivatives

y =x +1x2

y = (5ax2 )(12x!3)y = sin x

y = x5

y = a

y = xn

Some integrals

xn dx!

dx!

xdx!adx!

(ay3 ± by2 )! dy

(v0 + at)dt!

cos! d!"

Chapter 3���Kinematics in two and three dimensions

45

x and y components

Start with 1D Motion

46

v = v0 + at

vavg =12(v0 + v)

x = x0 + vavt

x = x0 + v0t +12at 2

v2 = v02 + 2a(x ! x0 )

3 independent equations Derive these 2 from the other 3

Example A projectile at 1.5 m high is shot horizontally with speed v0 that causes it to land 1.25 m away. What is the time it is in the air What is v0? What is vx and vy and the speed v when it lands?

1.5 m

1.25 m

v0

x

y

x0

y0

0

47