PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 5:

PHY 113 A Fall 2012 -- Lecture 5 19/7/2012

PHY 113 A General Physics I9-9:50 AM MWF Olin 101

Plan for Lecture 5:

Chapter 4 – Motion in two dimensions

1. Position, velocity, and acceleration in two dimensions

2. Two dimensional motion with constant acceleration



In the previous lecture, we introduced the abstract notion of a vector. In this lecture, we will use that notion to describe position, velocity, and acceleration vectors in two dimensions.

iclicker exercise:

Why spend time studying two dimensions when the world as we know it is three dimensions?

A. Because it is difficult to draw 3 dimensions.B. Because in physics class, 2 dimensions are

hard enough to understand.C. Because if we understand 2 dimensions,

extension of the ideas to 3 dimensions is trivial.

D. On Fridays, it is good to stick to a plane.


i

j vertical direction (up)

horizontaldirection

jir ˆ)(ˆ)()( tytxt


Vectors relevant to motion in two dimenstions

Displacement: r(t) = x(t) i + y(t) j

Velocity: v(t) = vx(t) i + vy(t) j

Acceleration: a(t) = ax(t) i + ay(t) j

dtdx

x vdtdy

y v

dt

dvxx a

dt

dvyy a


Visualization of the position vector r(t) of a particle

r(t1)r(t2)


Visualization of the velocity vector v(t) of a particle

r(t1)

r(t2)

12

12

0

)()(lim

12 tt

tt

dt

dt

tt

rrrv

v(t)


Visualization of the acceleration vector a(t) of a particle

r(t1)

r(t2)

12

12

0

)()(lim

12 tt

tt

dt

dt

tt

vvva

v(t1) v(t2)

a(t1)


Animation of position vector components associated with trajectory motion


Animation of velocity vector and its components associated with trajectory motion


Animation of acceleration vector associated with motion along a path


Projectile motion (near earth’s surface)

i

j vertical direction (up)

horizontaldirection

ja

jiv

jir

ˆ)(

ˆ)(ˆ)()(

ˆ)(ˆ)()(

gt

tvtvt

tytxt

yx



jv

a

jir

v

jir

ˆ)(

ˆ)(ˆ)()(

ˆ)(ˆ)()(

gdt

dt

tvtvdt

dt

tytxt

yx

ii tgtt

gdt

dt

vvjvv

jv

a

0 that note ˆ

ˆ)(



jv

a

jir

v

jir

ˆ)(

ˆ)(ˆ)()(

ˆ)(ˆ)()(

gdt

dt

tvtvdt

dt

tytxt

yx

iii

i

tgttt

gtdt

dt

rrjvrr

jvr

v

0 that note ˆ

ˆ)(

221



ˆ)( jvr

v gtdt

dt i

jv

a ˆ)( gdt

dt

jvrr ˆ221 gttt ii

iiyi

iixi

yixii

v

v

vv

sin

cos

ˆˆ

v

v

jiv


Projectile motion (near earth’s surface) Trajectory equation in vector form:

ˆ)( jvv gtt i jvrr ˆ221 gttt ii

Aside: The equations for position and velocity written in this way are call “parametric” equations. They are related to each other through the time parameter.

Trajectory equation in component form:

2

21 gttvyty

tvxtx

yii

xii

gtvtv

vtv

yiy

xix

)(

)(



Trajectory path y(x); eliminating t from the equations:

Trajectory equation in component form:

2

212

21 sin

cos

gttvygttvyty

tvxtvxtx

iiiyii

iiixii

gtvgtvtv

vvtv

iiyiy

iixix

sin)(

cos)(

2

21

2

21

costan

coscossin

cos

ii

iiii

ii

i

ii

iiii

ii

i

v

xxgxxyxy

v

xxg

v

xxvyxy

v

xxt


Projectile motion (near earth’s surface) Summary of results

2

21

221

costan

sin)( cos)(

sin cos

ii

iiii

iiyiix

iiiiii

v

xxgxxyxy

gtvtvvtv

gttvytytvxtx

iclicker exercise:

These equations are so beautiful thatA. They should be framed and put on the wall.B. They should be used to perfect my

tennis/golf/basketball/soccer technique.C. They are not that beautiful.


h=7.1mqi=53o

d=24m=x(2.2s)


Problem solving steps1. Visualize problem – labeling variables2. Determine which basic physical principle(s) apply3. Write down the appropriate equations using the variables defined in

step 1.4. Check whether you have the correct amount of information to solve the

problem (same number of known relationships and unknowns).5. Solve the equations.6. Check whether your answer makes sense (units, order of magnitude,

etc.).

h=7.1mqi=53o

d=24m=x(2.2s)


sms

mv

vxd

tvxtx

oi

oi

iii

/12698.182.253cos

24

242.253cos)2.2(

cos

h=7.1mqi=53o

d=24m=x(2.2s)

PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 5:

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