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Course Requirements

• 3 long exams (20% each) + 1 Final exam (25%)

• Quizzes/Problem sets/Recits (15%)

 All exams are Departmental!

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Chapter 2: Motion along a

Straight Line

Second Semester AY 2012-2013

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Recall:

2 1 

2 1

av x

 x x   xv

t t t 

 

2 1

 

2 1

 x x x

av x

v v v

a t t t 

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Motion with Constant

 Acceleration

• In terms of velocity, acceleration, and

time

• In terms of position, velocity,

acceleration, and time

• In terms of position, velocity, and

acceleration

• In terms of position, velocity, and time

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In Terms of Velocity,

 Acceleration, and TimeUsing:

2 1 

2 1

 x xav x

v va

t t 

2 1

2 1

 x x x

v va

t t 

If we let:

v1x=v0x at t1=0 and

v2x=vx at t2=t

We have:

0 x x x

v va

t 

0 x x xa t v v

Therefore:

0 x x xv v a t  

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In Terms of Position, Velocity,

 Acceleration, and TimeUsing:

2 1

 

2 1

av x

 x xv

t t 

If we let:

x1=x0 at t1=0 and

x2=x at t2=t

We have:

0 av x

 x xv

t 

Our expression

for the average

velocity

 An alternate

expression for the

average velocity:0

 2

 x xav x

v vv

 

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2

0 0

1

2 x x x x v t a t  0

1

2av x x xv v a t  

In Terms of Position, Velocity,

 Acceleration, and Time

0 av x

 x xv

t 

0

 2

 x xav x

v vv

 

0 x x xv v a t  

So far, we have:

1 2 3

3 2in

0 0

1

2av x x x xv v v a t  

4

1 4in

00

1

2 x x

 x xv a t 

t 

Therefore:

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In Terms of Position, Velocity,

and Acceleration

0 x x xv v a t  

Using:0 x x

 x

v vt 

a

2

0 0

1

2 x x x x v t a t 

Note:1 2

1 2in2

0 00 0

1

2

 x x x x x x

 x x

v v v v x x v a

a a

Therefore:

2 2

0 02 x x xv v a x x

By algebra… 

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In Terms of Position, Velocity,

and TimeUsing:

0 av x

 x xv

t 

and 0

 2

 x xav x

v vv

 

1 2

1 2&Equate

0 0

2

 x x x x v v

t 

Therefore:

00

2

 x xv v x x t 

 

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Summary

0 x x xv v a t  

2

0 0

1

2 x x x x v t a t 

2 2

0 02 x x xv v a x x

00

2

 x xv v x x t   

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 An airplane travels 280 m down a runway before taking off. If itstarts from rest, moves with constant acceleration, and becomes

airborne in 8.00 s, what is its speed, in m/s, when it takes off?

Example

00

0 0

2

: 0; 0

 x x

 x

v v

 x x t 

where x v

 

2

 xv

 x t 

 

2 280270.0

8 x

m x mv

t s s

So:

By algebra:

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 An antelope moving with constant acceleration covers the

distance between two points 70.0 m apart in 7.00 s. Its speed

as it passes the second point is 15.0 m/s.

a.) What is its speed at the first point?

0 0 

0

Using:2

 

 x xav x av x

 x

 x x v vv and v

t 

we equate them and solve for v

0

0

2 2 7015.0 5.00

7.00 x x

 x x m   m mv v

t s s s

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0 x x xv v a t  

0

2

15.0 5.00

1.437.00

 x x x

m mv v   m s sa

t s s

and the result of part a.)

 An antelope moving with constant acceleration covers the

distance between two points 70.0 m apart in 7.00 s. Its speed

as it passes the second point is 15.0 m/s.

b.) What is the acceleration?

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Problem 2.21 

b.) What is the acceleration? Alternate solution!!

and2

0 0

1

2 x x x x v t a t  Combine

0 x x xv v a t  

This will prevent us from using the result of part a.); it will also yield thesame result!

21.43 x

ma

 s

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Do you want to see

something cool?

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Thank you very much!See you all next meeting!