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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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