1
Unit 1
Motion in 1dim
Part 2
2
•
Graphical Analysis of 1dim
motion
•
Motion with Constant
Acceleration
•
Solving Problems with
a=const
•
Freely Falling Objects
Presentation Goals
CT #1:
A particle is moving in a
straight line that its position is given
by the equation
x(t) = 2.10 t
2
+ 2.80
The particle’s average acceleration
during the time interval
t
1
=3.0 s
and
t
2
=5.00
s is
2.10 m/s2
2.80 m/s2
4.20 m/s2
5.00 m/s2
0%
0%
0%
0%
Graphical Analysis of 1dim
1.
2.10 m/s
2
2.
2.80 m/s
2
3.
4.20 m/s
2
4.
5.00 m/s
2
0 of 200
10
CT #2:
A particle is moving in a
straight line that its position is
given by the equation
x(t) =2.10 t
2
+2.80
The particle’s instantaneous
acceleration at
t=5.00 s
is
2.10 m/s2
2.80 m/s2
4.20 m/s2
5.00 m/s2
0%
0%
0%
0%
Graphical Analysis of 1dim
1.
2.10 m/s
2
2.
2.80 m/s
2
3.
4.20 m/s
2
4.
5.00 m/s
2
0 of 200
10
5
•
Q:
Can we construct
an analytical
model
of
ANY
motion with
a=const
?
•
Q#1:
What is the velocity vs. time?
•
A:
for
any
interval
∆
t =tt
0
=t
0
•
dv/dt
= const = a,
so
•
Now, we can solve the velocity vs.
time
:
t
a
+
=
0
v
v
Constant Acceleration
(A)
at
v
v
dt
a
dv
v
v
t
=
−
⇒
=
∫
∫
0
0
0
6
•
Q:
How to represent
graphically this model?
Constant Acceleration
•
A:
Its
velocity vs time
graph is a
straight line,
but not necessarily through the origin
•
Q#2:
How the position of the particle varies
with time when we know
a=const
and the initial
conditions
x
0
and
v
0
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7
•
Fact:
For
any
motion with t
0
=0:
•
In addition, as the velocity is increasing at a
constant rate
, the average velocity is
arithmetic
average
of the velocity at start and at the end:
•
we can eliminate
in Eq. 22 and get the
displacement vs. time:
(B)
t
x
t
t
x
0
0
0
x
x
v
−
=
−
−
=
v
)
v
v
(
2
1
v
0
+
=
t
x
x
)
v
v
(
2
1
0
0
+
=
−
Constant Acceleration
(22)
(29)
8
•
Eliminating the final velocity from
v=v
0
+at
we
find another form of the displacement vs. time:
(C)
2
0
0
2
1
v
at
t
x
x
+
=
−
t
x
x
)
v
v
(
2
1
0
0
+
=
−
Constant Acceleration
•
or eliminating the initial velocity from
v
0
=vat
we find the third form:
2
0
2
1
v
at
t
x
x
−
=
−
(D)
(B)
9
(E)
•
Q#3:
How to calculate
v
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 Spring '09
 dziembowski
 Acceleration, Velocity, Alice

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