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Zero Acceleration

by doppelganger007
Tags: acceleration
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doppelganger007
#1
Oct21-07, 02:18 PM
P: 18
1. The problem statement, all variables and given/known data
Let c be a path in R^3 with zero acceleration. Prove that c is a straight line or a point.


2. Relevant equations
F(c(t)) = ma(t)
a(t) = c''(t)


3. The attempt at a solution
so i know that since the acceleration is zero, the velocity must be constant, and when you integrate a constant, you get a straight line...but how to I prove mathematically that the velocity is constant, because you can't integrate 0dt, as far as I know?
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arildno
#2
Oct21-07, 02:20 PM
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The indefinite integral, i.e, the anti-derivative of 0 is, indeed, a constant; that is we have:
[tex]\int{0}dx=C[/tex]
doppelganger007
#3
Oct21-07, 02:30 PM
P: 18
oh ok, so if I integrate that again I get that c(t) = Ct + D, which fits the general equation for a line

but then, does that also prove that c(t) could just be a single point?

arildno
#4
Oct21-07, 02:47 PM
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Zero Acceleration

Indeed, since big C could be..0!
doppelganger007
#5
Oct21-07, 03:00 PM
P: 18
oh. duh!
gratzie
HallsofIvy
#6
Oct21-07, 05:27 PM
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Quote Quote by doppelganger007 View Post
1. The problem statement, all variables and given/known data
Let c be a path in R^3 with zero acceleration. Prove that c is a straight line or a point.


2. Relevant equations
F(c(t)) = ma(t)
a(t) = c''(t)


3. The attempt at a solution
so i know that since the acceleration is zero, the velocity must be constant, and when you integrate a constant, you get a straight line...but how to I prove mathematically that the velocity is constant, because you can't integrate 0dt, as far as I know?
Damn, I hate mixed "physics" and "mathematics" problems! You or whoever set this problem, should know that a "path" DOES NOT HAVE an "acceleration". I expect this problem should be "find the equation of motion of a particle whose trajectory is a given path in R3 with acceleration 0. Show that the path is either a straight line or a point". Then you would begin with [itex]\vec{a}= d\vec{v}/dt=[/itex] and go from there.
Gib Z
#7
Oct22-07, 05:42 AM
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The easiest way would have been to recognise that acceleration is a vector quantity, it is affected both by direction or magnitude. No acceleration, no change in direction, which means constant gradient. Simple as that.


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