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

In summary, the problem asks for the equation of motion of a particle that has a zero acceleration, which is not possible. Therefore, the solution is to change the problem to find the equation of motion for a particle that has a given acceleration.
  • #1
doppelganger007
18
0

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?
 
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  • #2
The indefinite integral, i.e, the anti-derivative of 0 is, indeed, a constant; that is we have:
[tex]\int{0}dx=C[/tex]
 
  • #3
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?
 
  • #4
Indeed, since big C could be..0!
 
  • #5
oh. duh!
gratzie
 
  • #6
doppelganger007 said:

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 [itex]\vec{a}= d\vec{v}/dt=[/itex] and go from there.
 
Last edited by a moderator:
  • #7
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.
 

1. What is the definition of a straight line?

A straight line is a geometric figure that extends infinitely in both directions and has a constant slope (or rate of change) between any two points on the line. It can also be described as the shortest distance between two points.

2. How is a straight line represented mathematically?

A straight line can be represented mathematically using the slope-intercept form, y = mx + b, where m is the slope of the line and b is the y-intercept (the point where the line intersects the y-axis).

3. What does it mean for a function to have zero acceleration?

When a function has zero acceleration, it means that its rate of change or velocity is constant. In other words, the function is not speeding up or slowing down, it is moving at a constant speed.

4. How do you prove that a function is a straight line?

To prove that a function is a straight line, you can use the slope formula (m = (y2-y1)/(x2-x1)) to calculate the slope between any two points on the line. If the slope is constant (meaning it is the same for any two points), then the function is a straight line.

5. What is the process for proving that a function has zero acceleration?

The process for proving that a function has zero acceleration involves calculating the derivative of the function twice. If the second derivative (or the rate of change of the first derivative) is equal to zero, then the function has zero acceleration and is a straight line.

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