Question 2: derive 0 = 1/2at2 + wot Oo- You may use prior derived results.

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Chapter1: Units, Trigonometry. And Vectors
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The Notion of Deriving
Deriving an equation (or a statement) is more than just asserting it. It's
showing how it must be true (and anything contrary must be false).
One doesn't understand the material as long as one is limited to
memorizing statements or equations, without being able to connect
them.
As a further, practical thing, understanding the connection makes it less
likely that one would use an equation when it doesn't apply. In
particular, one would avoid the "equations of constant acceleration"
anytime one had non-constant acceleration.
In this assignment, I assign you to derive the "equations of constant
acceleration" -- all but one that I will derive. We will use 0, w, and a,
satisfying these definitions:
dw
de
and
a =
dt
W =
dt
and one special case: a is constant. (You should spot the false step
when a is non-constant.) w and 0 are functions of time: w(t) and 0(t).
ωο- ω(0), and θ0-θ(0).
My Example: derive w2 - wo? = 2a(0-0).
dw
dw de
dw
a =
dt
de dt
de
Multiply by dt and integrate:
Sado
w dw = w?
d0 = a0 + C =
w?
2a0 + 2C
Substitute the initial (t=0) quantities, and subtract. The constant of
integration vanishes.
w3 = 2a0 + 2C
w? – w = 2a(0 – 0o)
The e integration fails if a is non-constant. There is also a type of
situation where the step before the integration fails, involving w.
Transcribed Image Text:The Notion of Deriving Deriving an equation (or a statement) is more than just asserting it. It's showing how it must be true (and anything contrary must be false). One doesn't understand the material as long as one is limited to memorizing statements or equations, without being able to connect them. As a further, practical thing, understanding the connection makes it less likely that one would use an equation when it doesn't apply. In particular, one would avoid the "equations of constant acceleration" anytime one had non-constant acceleration. In this assignment, I assign you to derive the "equations of constant acceleration" -- all but one that I will derive. We will use 0, w, and a, satisfying these definitions: dw de and a = dt W = dt and one special case: a is constant. (You should spot the false step when a is non-constant.) w and 0 are functions of time: w(t) and 0(t). ωο- ω(0), and θ0-θ(0). My Example: derive w2 - wo? = 2a(0-0). dw dw de dw a = dt de dt de Multiply by dt and integrate: Sado w dw = w? d0 = a0 + C = w? 2a0 + 2C Substitute the initial (t=0) quantities, and subtract. The constant of integration vanishes. w3 = 2a0 + 2C w? – w = 2a(0 – 0o) The e integration fails if a is non-constant. There is also a type of situation where the step before the integration fails, involving w.
Question 2: derive 0 = 1/2at² + wot
+ 0o-
You may use prior derived results.
Transcribed Image Text:Question 2: derive 0 = 1/2at² + wot + 0o- You may use prior derived results.
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