Physics 40 Chapter 21-D Kinematics

Natural Motion•Objects have a proper place •Objects seek their natural place•The heavier the object,

the faster it falls.•Did not experiment to test theories.

The natural motion of a body is to remain in

whatever state of motion it is in unless

acted upon by net external forces.

Galileo Challenged The DogmaOf Natural Motion

Galileo Challenged The DogmaOf Natural Motion with

Experiments

Galileo Challenged Aristotle PhysicsIn a vacuum, all objects fall with the same

acceleration due to gravity: 9.80 m/s2, independent of their weight.

Galileo’s Motion Studies

What causes a rolling ball to stop?

What causes the ball to speed up?

What causes the ball to slow down?

gravity

gravity

friction

Galileo’s Motion Studies

0

2

f

xvt

v vv

vat

Δ=Δ+

=

Δ=Δ

gave us…

Definitions:

From these we derive the kinematics equations….

0 , , 2

fv vx vv v at t

+Δ Δ= = =Δ Δ

0

2fv vx

t+Δ

=Δ 0

2

f

t xv v

→Δ = Δ+

vat

Δ=Δ

Start:

Combine & Eliminate t:

0 = fv vt

a−

→ Δ

2 0

0

v vft x

v v af

−Δ = Δ =

+

2 20 2fv v a x= + ΔAlgebra:

Galileo’s Motion Studies

0

20

0

2 20

, , 2

12

2

f

f

f

v vx vv v at t

x v t at

v v at

v v a x

+Δ Δ= = =Δ Δ

Δ = +

= +

= + Δ

gave us…

Kinematic Equations

Newton’s Calculus

2

2

( ) ( ) ( )( ) , ( )

dx t dv t d x tv t a tdt dt dt

= = =

0

( )t

xx t v dt= ∫

0

( ) ( )t

v t a t dt= ∫

gave us…

Uniform Motion in a nutshellMotion with constant acceleration.

0Definitions: , , 2

fv vx vv v at t

+Δ Δ= = =Δ Δ

2

2

20

2 20 0

Kinematics Eqs:

, ,

1 2

, 2f f

dx dv d xv a adt dt dt

x v t at

v v at v v a x

= = =

Δ = +

= + = + Δ

Distance and Displacement

(delta) means "change in" = 'final - initial'Δ

Δ

0f

The total distance traveled relative to an origin.Distance is a scalar.

Displacement is a vector. The unit is the meter.

x x xΔ = −

Average Speed &Velocity Speed is how fast something moves.

The average speed is the total distance per time. The average velocity is the the total displacement per time.

Velocity is a vector. The unit is m/s.

total displacementtotal timeave

xvt

Δ= =

Δ

AccelerationHow fast How fast is changing.

The rate at which the speed is changing.

Speeding up

Slowing down

Constant speed, changing direction.

change in velocitychange in time

vat

Δ= =

Δ

Acceleration is in the direction of the net Force but not necessarily

in the direction of velocity.Velocity is always in the direction of the motion!

Sense of Speed

1 / 3.6 / 2.24 /10 / 36 / 22.4 /20 / 72 / 44.8 /30 / 108 / 67.2 /

m s km hr mi hrm s km hr mi hrm s km hr mi hrm s km hr mi hr

= == == == =

1 / 2.25 /m s mi hr≈

1 / 0.62 /km hr mi hr≈

Quicky Question

An automobile enters a freeway on-ramp at 15.0m/s and accelerates uniformly up to 25.0 m/s in a time of 10.0s. a) What is the automobile’s average velocity?

0

2f

ave

v vv

+=

15 / 25 / 20 /2

m s m s m s+= =

Which equation?

An automobile enters a freeway on-ramp at 15.0m/s and accelerates uniformly up to 25.0 m/s in a time of 10.0s.

b) What is the automobile’s average acceleration?

f iv vvat t

−Δ= =Δ Δ

225 / 15 / 1 /10

m s m s m ss−

= =

Which equation?

Quicky Question

An automobile enters a freeway on-ramp at 15.0m/s and accelerates uniformly up to 25.0 m/s in a time of 10.0s.

c) What is the distance traveled in this amount of time?

Which equation?

Quicky Question

20

1

2x v t atΔ = + 2

2

1

215 / (10 ) 1 (10 )mm s s s

s= +

200x mΔ =

21 /a m s=

(you could also use vave equation.)

Acceleration: Changing Velocity QUESTION:

From t = 0, how long does it take the car to come to a full stop?

+x

YOU TRY IT!

Acceleration: Changing Velocity

2

Knowns5 /

28 /0

?

i

f

a m sv m sv

t

= −==

=

f iv vt

a−

=

2

0 2 8 / 5 .65 /

m s sm s

−= =

−

f iv v at= +Which equation to use?

Solve for t:

5.6t s=

Acceleration: Changing Velocity From t = 0, to t = 5.6s, how far does the car travel before it

comes to a stop?

+x

2

Knowns5 /

28 /0

5.6

i

f

a m sv m svt s

= −==

=

Which equation? 20

12

x v t atΔ = +

2 2128 5.6 ( 5 / )(5.6 ) 78.42

mx s m s s ms

Δ = + − =

78.4x mΔ =

YOU TRY IT!

What Goes Up Must Come DownSomeone standing at the edge of a cliff throws one ball straight up and one straight down at the same speed. Ignoring air resistance, which ball strikes the ground with the greatest speed?

Brake QuestionYou are driving a car going 80 km/hr (50mph) when a head-on collision happens 25 meters ahead of you. If you can brake at 6 m/s2, will you hit the crash or stop before it?

2 20 2fv v a x= + Δ

20

2vx

aΔ = −

2

2

(22 / ) 40.3 252( 6 / )

m sx m mm s

Δ = − = >−

20: 80 / 22 / , 0, 6 / fKnowns v km hr m s v a m s= = = = −

: ?Unknown xΔ =

CRASH!

Stopping Distance goes as the SQUARE of the speed!

Free FallUnless told otherwise, ignore air resistance for

free fall problems!

Acceleration of Freely Falling Object

• The acceleration of an object in free fall is directed downward, regardless of the initial motion

• The magnitude of free fall acceleration is g = 9.80 m/s2

– g decreases with increasing altitude– g varies with latitude– 9.80 m/s2 is the average at the Earth’s surface– We will neglect air resistance

Free Fall EquationsFor any object in the absence of air resistance.

29.80 /ya g m s= − = −

0

20

2 20

Customize:

12

2

f

f

v v gt

y v t gt

v v g y

= −

Δ = −

= − Δ

0

20

2 20

Kinematic Eqs:

1 22

f

f

v v at

x v t at

v v a x

= +

Δ = +

= + ΔNote: v0 can be negative!

(taking up as +y)

Falling from Rest

2 21 52

y at tΔ = =

2

:~ 10 /

Estimatea g m s=

10v at t= =

20 /20

v m sy m=

Δ =

10 /5

v m sy m=

Δ =

30 /45

v m sy m=

Δ =

40 /80

v m sy m=

Δ =

50 /125

v m sy m=

Δ =

+

0

20

12

fv v gt

y v t gt

= +

Δ = +

0 0v =

!v y≠ ΔHow FAR is notHow FAST!

Take down as +y:

How Far: y(t) ~ t2

0fv v at= +

20

12

y v t atΔ = +

How Fast: v(t) ~ t1

+

Free FallQuestion: You throw the rock down with an initial speed of 30 m/s. The rock hits the ground in 3 seconds. With what velocity will the rock hit the ground?

+y

230 9.8 (3 )m m ss s

= − −

59.4fmvs

= −

0fv v at= +

How high is the cliff?

20: 30 / , 9.8 / , 3Knowns v m s a m s t s= − = − =

: ?fUnknown v =

Free Fall

20

1

2y v t atΔ = +

2 21

2( 30 / )(3 ) ( 9.8 / )(3 )m s s m s s= − + −

134m= −

The cliff is 134 m high.

20: 30 / , 9.8 / , 3Knowns v m s a m s t s= − = − =

: ?Unknown yΔ =+y

Question: You throw the rock down with an initial speed of 30 m/s. The rock hits the ground in 3 seconds. With what speed will the rock hit the ground? How high is the cliff?

Free Fall: Throwing UpWhat is the speed at the top of the path?ZERO!

What is the acceleration at the top?a = -9.80 m/s2

What is the velocity at the same height on the way down?-30 m/s

+y

With what velocity will the rock hit the ground?-59.4 m/s SAME as if you threw it straight down at 30m/s!

How long does it take to hit the ground? First try to guess!

+y0fv v at= +

02

59.4 / 30 /9.8 /

fv v m s m sta m s− − −

= =−

9.12t s=

20: 30 / , 9.8 / , 3 , 59.4 /fKnowns v m s a m s t s v m s= = − = = −

: ?Unknown tΔ =

How long to the top? How long back to launch point? Final v increases by 30m/s?

I guess about 9 seconds!

Free Fall: Throwing Up Problem

Symmetry of The G-Field

Motion Graphs

What kind of motion is this?

xvt

Δ=Δ

3400 1 /400

mv m ss

−= = −2 0 /v m s=

1400 2 /200

mv m ss

= =

What is the velocity during each segment?

Average VelocityWhat is the average velocity between B and D?

2( ) 4 2x t t t= − +

xxvt

Δ=Δ

6 ( 2 )2− −

=m m

s

(3 ) (1 )(3 1 )

−=

−x s x s

s s

4=ms

Instantaneous VelocityThe velocity at any time t is the slope of the x vs t graph at t.

( )xdxv tdt

=

2( 4 2 )( ) 4 4d t tv t tdt

− += = − +

2(2.5 ) 4 4 (2.5 ) 6m m mv s ss s s

= − + =

What is the instantaneous velocity at t=2.5s?

What kind of motion does this graph represent?

What does the velocity vs time graph look like?

2( ) 4 2x t t t= − +

Velocity Graph

2( ) 4 2x t t t= − +

What does the a-t graph look like?

Acceleration Graph

2( ) 4 2x t t t= − +

What does the a-t graph look like?

24 /xa m s=

xx

dvadt

=2

2

d xdt

=2 2

2

( 4 2 )d t tdt

− += 24 /m s=

What is the displacement from zero to 2s?

m/s

(s)

In general……

0

( )t

xx t v dt= ∫

0

( )t

xa t dt= ∫

Displacement = area under the v-t graph

1 ( )2 xa t t= 1 (base)(height)

2=

Area under graph=

212 xa t=

What is the displacement from zero to 2s?

m/s

(s)

1 (base)(height)2

x =

1 1(1 )(-4 / ) (1 )(4 / ) 02 2

= + =s m s s m s

What is the displacement from zero to 2s?2( ) 4 2x t t t= − +

Displacement = area under the v-t graph

(2 ) 0x s =

1 (base)(height)2

x =

1 1(1s)(-4m/s)+ (1s)(4m/s)=02 2

=

Zero!

m/s

(s)

What is the displacement from zero to 4s?2( ) 4 2x t t t= − +

Displacement = area under the v-t graph

(4 ) 16x s m=

1 (base)(height)2

x =

1 1(1s)(-4m/s)+ (3s)(12m/s)=16m2 2

=

m/s

(s)

HW Problem #46

Derive displacement equations for each segment.

HW # 60 Rock Drop

A rock is dropped from rest into a well. The sound of the splash is heard 2.40 s after the rock is released from rest. How far below the top of the well is the surface of the water? The speed of sound in air (at the ambient temperature) is 336 m/s. 

HW #32 Speedy SueSpeedy Sue, driving at 30.0 m/s, enters a one‐lane tunnel.  She then observes a slow‐moving van 155 m ahead traveling at 5.00 m/s.  Sue applies her brakes but can accelerate only at −2.00 m/s2 because the road is wet.   Will there be a collision?  If yes, determine how far into the tunnel and at what time the collision occurs.  If no, determine the distance of closest approach between Sueʹs car and the van.

Hill Question

On which of these hills does the ball roll down with increasing speed and decreasing acceleration? a) b) c)

a) b) c)