Wiley Plus

Assignment 1 is online:

6 problems from chapters 2 and 31D and 2D Kinematics

Due Monday October 5Before 11 pm !

Chapter II: Kinematics In One Dimension

• Displacement

• Speed and Velocity

• Acceleration

• Equations of Kinematics for Constant Acceleration

• Applications of the Equations of Kinematics

• Freely Falling Bodies

• Graphical Analysis of Velocity and Acceleration

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Graphical Analysis of Velocity and Acceleration

Average speed

The slope of the curve is constant, so the speed is constant.

m/s42/8/ txv

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m/s2200

4001 v

m/s0400

02 v m/s1

400

4003

v

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Instantaneous speed at = 20 s is (26 m)/(5 s) = 5.2 m/s

Speed is not constant

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Acceleration = (12 m/s)/(2 s) = 6 m/s2

The slope of the curve is constant, so the acceleration is constant

Constant Acceleration

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Example: A person who walks for exercise produces the position-timegraph given below.

a) Without any calculations, decide which segment of the graph (A, B, C, or D) indicates a negative average velocity.

B) decide which segment indicates a zero average velocity.

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Example: A bus makes a trip according to the position-time graph shown below.

What is the average acceleration (in km/h2) of the bus for the entire 3.5 hour period?

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Chapter 3

• Kinematics in Two Dimensions

• Equations of Kinematics in Two Dimensions

• Projectile Motion

• Displacement, velocity, acceleration extended totwo dimensions

• Motion in x can be separated completely frommotion in y, provided air resistance is negligible – treatment of projectile motion

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o

o

ttxx

time elapsedDistance

speed Average

otttime elapsedntDisplaceme

velocity Average

x

t Velocity ousInstantane

t

xv

0lim

otttime elapsedvelocity in change

onaccelerati Average

ovv

tonAccelerati ousInstantane

t

v

0lim

Speed, Velocity and Accelerationin One Dimension

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0rrr

ntDisplaceme

ottvelocity Average

r

t Velocity ousInstantane

t

r

0lim

ottonaccelerati Average

ovv

tonAccelerati ousInstantane

t

v

0lim

0v

v

Position vectors r, r0 at t, t0

There is an acceleration whenever there is a change of speed or direction !

Speed, Velocity and AccelerationIn Two Dimensions

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Vectors can be resolved into components

yv

xv

y

x

ˆsin

ˆcos

v

v

Instantaneous velocity v

The components separately followthe same equations of motion as in the one dimensional case, since the motion for each component happens in one dimension !

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tavv xxx 0

2

002

1tatvxx xx

tvvxx xx 002

1

2

0

22 xxox vvxxa

Equations of Kinematics in Two Dimensions

1)

2)

3)

4)

tavv yyy 0

2

002

1tatvyy yy

tvvyy yy 002

1

2

0

22 yyoy vvyya

Same as before, only with subscripts for each direction of motion

x y

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Problem 3.8: A skateboarder rolls down a 12 m ramp, reaching a speed of 7.7 m/s at the bottom.

What is her average acceleration?If the ramp has an angle of 25 degrees with respect to the horizontal, what is the component of acceleration in the horizontal direction?

x

y

00v

m/sv 77.

m12

a

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x

y

00v

m/sv 77.

m12

a

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Example:

A spacecraft is traveling with a velocity of v0x = 5480 m/s along the positive x direction. Two engines are fired for 842 seconds.

What is the final speed in the x and y directions?What is the total final velocity?

2y

2x

m/sa :2 Engine

m/sa :1 Engine

48

21

.

.

yv

xv

v

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yv

xv

v

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Projectile Motion

• Consider motion in x and y separately

• Ignore air resistance velocity in x-direction is constant

• Write down positions in x and y as a function of time

• Remember that the projectile travels vertically (up and down – y)in the same time that it is traveling above the horizontal (x)

• The only acceleration is that due to gravity, acting downward(a rocket or an object which is self propelled is not considered aprojectile and does not undergo projectile motion, because it can be accelerated arbitrarily in any direction.)

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In the absence of air resistance:no forces act in x-direction, so vx, the speed in x-direction is constant throughout the path.

Speed changes in y-direction because of gravity.

g

x

yconstant isvx

yygayˆ.ˆ

289

sm

g0xa

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tvx x0 tvy y0

Projectile motion therefore follows that of a parabola:

2

2

1tax 2

2

1tay

2

2

1gt

tvx x0

xv

xt

0

tvy y0 2

2

1gt

2

00

0

2

1x

v

gx

v

vy

xx

y

Equation of an upside downparabola in x and y

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