physics 207: lecture 3, pg 1 physics 207, lecture 3 l today (finish ch. 2 & start ch. 3) ...

Physics 207: Lecture 3, Pg 1

Physics 207, Lecture 3 Today (Finish Ch. 2 & start Ch. 3)

Examine systems with non-zero acceleration (often constant)

Solve 1D problems with zero and constant acceleration (including free-fall and motion on an incline)

Use Cartesian and polar coordinate systems

Perform vector algebra


AccelerationAcceleration

The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval in which that change occurs.

Bold fonts are vectors

a The average

acceleration is a vector quantity directed along ∆v


Average AccelerationAverage Acceleration

Question: A sprinter is running around a track. 10 seconds after he starts he is running north at 10 m/s. At 20 seconds he is running at 10 m/s eastwards.

What was his average acceleration in this time (10 to 20 secs)?

t

vv

tt

tvtva if

if

if

initialfinalavg

)()(


Average AccelerationAverage Acceleration

Question: A sprinter is running around a track. 10 seconds after he starts he is running north at 10 m/s. At 20 seconds he is running at 10 m/s eastwards.

What was his average acceleration in this time (10 to 20 secs)?

t

vv

tt

tvtva if

if

if

initialfinalavg

)()(

vvv if

14 m/s /10 s= 1.4 m/s2 to the SE

N


Instantaneous Acceleration

Average acceleration

The instantaneous acceleration is the limit of the average acceleration as ∆v/∆t approaches zero

t

vva if

avg


Position, velocity & acceleration for motion along a line

If the position x is known as a function of time, then we can find both the instantaneous velocity vx and instantaneous acceleration ax as a function of time!

t

vx

t

x

ax

t

2

2va

dtxd

dtd x

x

dtdx

x v

] offunction a is [ )( txtxx


Going the other way…. Particle motion with constant acceleration

The magnitude of the velocity vector changesxa

dtd x

x

va

xx ddt va

ixxxx t v-vva

a

ti

t0

t

tf

v = area under curve = a t


Going the other way…. Particle motion with constant acceleration

The magnitude of the velocity vector changes A particle with smoothly increasing speed:

vv11vv00 vv33 vv55vv22 vv44

aa

v

t0

vf = vi + a t

= vi + a (tf - ti )

v = area under curve = a t

xa

a

ti

t0

t

tf

txxxi

avv


So if constant acceleration we can integrate twice

txxxi

avv2

21 a v ttxx xxi

i

consta xt

ax

x

t

xi

vx

t

vi


Two other relationships

If constant acceleration then we also get:

)( 2avv 22ixxx xx

i

)vv(v 21

(avg) xxxi


Example problem

A particle moves to the right first for 2 seconds at 1 m/s and then 4 seconds at 2 m/s.

What was the average velocity?

Two legs with constant velocity but ….

vx

t

2

vvv 21

Avg


Example problem

A particle moves to the right first for 2 seconds at 1 m/s and then 4 seconds at 2 m/s.

What was the average velocity?

Two legs with constant velocity but ….

We must find the displacement (x2 –x0) And x1 = x0 + v0 (t1-t0) x2 = x1 + v1 (t2-t1) Displacement is (x2 - x1) + (x1 – x0) = v1 (t2-t1) + v0 (t1-t0) x2 –x0 = 1 m/s (2 s) + 2 m/s (4 s) = 10 m in 6 seconds or 5/3 m/s

vx

t

2vv

v 21Avg


A particle starting at rest & moving along a line with constant acceleration has a displacement

whose magnitude is proportional to t2

221 a)( txx xi

221 a v ttxx xxi

i

1. This can be tested2. This is a potentially useful result

Displacement with constant acceleration


Free Fall

When any object is let go it falls toward the ground !! The force that causes the objects to fall is called gravity.

This acceleration on the Earth’s surface, caused by gravity, is typically written as “little” g

Any object, be it a baseball or an elephant, experiences the same acceleration (g) when it is dropped, thrown, spit, or hurled, i.e. g is a constant.

221

0 g v)(0

ttyty y ga -y


Gravity facts:

g does not depend on the nature of the material !

Galileo (1564-1642) figured this out without fancy clocks & rulers!

Feather & penny behave just the same in

vacuum

Nominally, g = 9.81 m/s2 At the equator g = 9.78 m/s2

At the North pole g = 9.83 m/s2


When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path?

A. Both v = 0 and a = 0

B. v 0, but a = 0

C. v = 0, but a 0

D. None of the above

y

Exercise 1Motion in One Dimension


When throwing a ball straight up, which of the following is When throwing a ball straight up, which of the following is true about its velocity true about its velocity vv and its acceleration and its acceleration aa at the highest at the highest point in its path?point in its path?

A. Both v = 0 and a = 0

B. v 0, but a = 0

C. v = 0, but a 0

D. None of the above

y

Exercise 1Motion in One Dimension


Exercise 2 1D Freefall

A. vA < vB

B. vA = vB

C. vA > vB

Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v0, Alice straight down and Bill straight up.

The speed of the balls when they hit the ground are vA and vB respectively.. ( (Neglect air resistance.)

vv00

vv00

BillAliceAlice

HH

vvAA vvBB


Exercise 2 1D Freefall

A. vA < vB

B. vA = vB

C. vA > vB

Alice and Bill are standing at the top of a cliff of heightAlice and Bill are standing at the top of a cliff of height HH. Both throw a ball with initial speed. Both throw a ball with initial speed vv00, Alice straight, Alice straight downdown and Bill straightand Bill straight upup. The speed of the balls when . The speed of the balls when they hit the ground arethey hit the ground are vvAA andand vvBB respectivelyrespectively..

v0

v0

BillAlice

H

vA vB


Problem Solution Method:

Five Steps:

1) Focus the Problem- draw a picture – what are we asking for?

2) Describe the physics- what physics ideas are applicable

- what are the relevant variables known and unknown

3) Plan the solution- what are the relevant physics equations

4) Execute the plan- solve in terms of variables

- solve in terms of numbers

5) Evaluate the answer- are the dimensions and units correct?

- do the numbers make sense?


A science project

You drop a bus off the Willis Tower (442 m above the side walk). It so happens that Superman flies by at the same instant you release the car. Superman is flying down at 35 m/s.

How fast is the bus going when it catches up to Superman?


A “science” project

You drop a bus off the Willis Tower (442 m above the side walk). It so happens that Superman flies by at the same instant you release the car. Superman is flying down at 35 m/s.

How fast is the bus going when it catches up to Superman?

Draw a picturey

t

yi

0


A “science” project

Draw a picture Curves intersect at

two points

y

t

yi

0

ttg

y Superman2 v

2

Supermanv 2

tg

Supermanv2g

t

SupermanSupermanbus v2v2

v g

gtg


See you Wednesday

Assignment: Assignment:

For Wednesday, Read through Chapter 4.4For Wednesday, Read through Chapter 4.4

physics 207: lecture 3, pg 1 physics 207, lecture 3 l today (finish ch. 2 & start ch. 3) ...

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