Prove c(t) is a Straight Line or Point w/ Zero Acceleration

[doppelganger007]

Homework Statement

Let c be a path in R^3 with zero acceleration. Prove that c is a straight line or a point.

Homework Equations

F(c(t)) = ma(t)
a(t) = c''(t)

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?

 

[arildno]

The indefinite integral, i.e, the anti-derivative of 0 is, indeed, a constant; that is we have:
\int{0}dx=C

 

[doppelganger007]

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]

Indeed, since big C could be..0!

 

[doppelganger007]

oh. duh!
gratzie

 

[HallsofIvy]

Homework Statement

Let c be a path in R^3 with zero acceleration. Prove that c is a straight line or a point.

Homework Equations

F(c(t)) = ma(t)
a(t) = c''(t)

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 \vec{a}= d\vec{v}/dt= and go from there.

 

[Gib Z]

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.