Physics 41
Lecture 2
Lecture 2: Uniform Acceleration in One Dimension
Goals of this Lecture
∙
To introduce some useful relationships between position, velocity, acceleration and time for
the special case of uniform (constant) acceleration.
2.1 Examples of Uniformly Accelerated Motion
During this course we will encounter many situations in which a particle is experiencing uniform
(constant) acceleration.
For example, a falling object has a velocity that steadily increases with
time because it is being accelerated uniformly by the force of gravity (see next lecture).
Even if an object has an acceleration that is varying with time,
we can always identify a small enough time interval,
u1D6FFu1D461
, over
which the acceleration is approximately constant, as shown in
the figure opposite. Consequently the results we derive here will
be very useful at several points in future lectures.
Note that Worked Example 4 from the previous lec
ture showed the velocity as a function of time for a
particular particle that is shown opposite.
This par
ticle is experiencing uniform acceleration.
You may
find it helpful to review that example as the method
used in this lecture to derive general expressions for
the position and velocity of a particle under constant
acceleration is exactly the same.
The procedure we will follow is (i) integrate the acceleration to obtain the velocity, then (ii) integrate
the velocity to obtain the position.
uni222B.alt01
u1D44Eu1D451u1D461
=
uni222B.alt01
u1D451u1D463
uni222B.alt01
u1D463u1D451u1D461
=
uni222B.alt01
u1D451u1D465
u1D44E
(
u1D461
)
=
⇒
u1D463
(
u1D461
)
=
⇒
u1D465
(
u1D461
)
2.2 Velocity as a Function of Time for a Uniformly Accelerating Particle
To see how to derive the velocity as a function of time for a particle that is experiencing uniform
acceleration, let’s solve the following problem:
Worked Example 1:
A particle is experiencing uniform acceleration as shown below left. If its
velocity at
u1D461
=
u1D461
u1D456
is
u1D463
u1D456
, find its velocity,
u1D463
u1D453
, at a time
u1D461
u1D453
=
u1D461
u1D456
+ Δ
u1D461
.
∙
Start with
u1D44E
=
u1D451u1D463
u1D451u1D461
and rearrange to obtain
u1D451u1D463
=
u1D44Eu1D451u1D461
∙
Integrate both sides from the initial time
u1D461
=
u1D461
u1D456
to the final
time
u1D461
=
u1D461
u1D453
:
u1D463
u1D453
uni222B.alt02
u1D463
u1D456
u1D451u1D463
=
u1D461
u1D453
uni222B.alt02
u1D461
u1D456
u1D44Eu1D451u1D461
(2.1)
∙
The left hand side is just
u1D463
u1D453
−
u1D463
u1D456
.
∙
The right hand side is the area under the curve which is
u1D44E
Δ
u1D461
1
Last updated December 30, 2009
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Physics 41
Lecture 2
∙
Now rearrange to the desired expression:
u1D463
u1D453
=
u1D463
u1D456
+
u1D44E
Δ
u1D461
(2.2)
We could also have evaluated the integral of the acceleration directly.
Since
u1D44E
is constant the
integral in Equation 2.1 becomes:
u1D463
u1D453
−
u1D463
u1D456
=
u1D44E
u1D461
u1D453
uni222B.alt02
u1D461
u1D456
u1D451u1D461
=
u1D44E
(
u1D461
u1D453
−
u1D461
u1D456
) =
u1D44E
Δ
u1D461
which gives the same result as before.
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 Winter '08
 Susskind,L
 mechanics, Acceleration, Velocity, 10 m, 0 m, 0 m/s, 6.0 m/s

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