In a constant acceleration motion, Is it possible to have constant speed?

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Homework Help Overview

The discussion revolves around the concept of constant acceleration in motion and whether it is possible to maintain a constant speed under such conditions. Participants explore the implications of acceleration on speed and the nature of motion in three dimensions.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants present differing views on the relationship between constant acceleration and constant speed, with some asserting that constant speed cannot occur with constant acceleration. Others introduce the idea of zero acceleration and question the definitions being used.

Discussion Status

The discussion is active, with participants questioning assumptions about acceleration and speed. Some have provided mathematical reasoning to support their points, while others are exploring the implications of different scenarios, such as uniform circular motion and the nature of vectors in three-dimensional space.

Contextual Notes

There is an ongoing debate about the definitions of acceleration and speed, particularly regarding the case of zero acceleration and its implications for motion. Participants are also considering the geometric interpretation of motion in three dimensions.

MatinSAR
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Homework Statement
In a constant acceleration motion, Is it possible to have constant speed?
Relevant Equations
Please see below.
I think it's not possible.
In 3D for constant acceleration we have : ##\vec v = \vec v_0 + \vec a t##

It's a line in 3 dimension so velocity's magnitude(speed) is changing with time.
I appreciate any better idea.
 
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You are correct, it is not possible. The closest you can come is constant speed and constant magnitude of the acceleration which is the case of uniform circular motion.
 
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##\vec a=0##?
 
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Ibix said:
##\vec a=0##?
Then we have no acceleration.
 
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MatinSAR said:
Then we have no acceleration.
Zero is a constant. If you mean "constant non-zero acceleration" I would say you should say so.
 
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MatinSAR said:
Then we have no acceleration.
That is true but ##0## is a perfectly good constant.
 
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Ibix said:
Zero is a constant. If you mean "constant non-zero acceleration" I would say you should say so.
You are right. I meant "constant non-zero acceleration". Thanks for your reply @Ibix .
kuruman said:
That is true but ##0## is a perfectly good constant.
Thank you for your time @kuruman .
 
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MatinSAR said:
It's a line in 3 dimension
Not sure what you mean. The trajectory from that equation does not have to be a straight line.
But how about a proof?
Constant speed implies ##\vec v\cdot\vec v=c##. Differentiating, ##\vec v.{\vec a}=0=\vec v_0\cdot\vec a+a^2t##. That can only be true for varying t and constant ##\vec a## if a=0.
 
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haruspex said:
Not sure what you mean. The trajectory from that equation does not have to be a straight line.
Why? Isn't it similar to ##\vec r=\vec r_0 + s\vec t## which is equation of a line in vector form?
haruspex said:
But how about a proof?
Constant speed implies ##\vec v\cdot\vec v=c##. Differentiating, ##\vec v.{\vec a}=0=\vec v_0\cdot\vec a+a^2t##. That can only be true for varying t and constant ##\vec a## if a=0.
Far better idea!
Thanks a lot for your time.
 
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  • #10
MatinSAR said:
Why? Isn't it similar to ##\vec r=\vec r_0 + s\vec t## which is equation of a line in vector form?
Consider ##\vec a## normal to ##\vec v_0##.
 
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  • #11
haruspex said:
Consider ##\vec a## normal to ##\vec v_0##.
Good point, Thanks! I didn't think about it.
 
  • #12
The equations used to describe motion with constant acceleration include, as a possibility, a constant acceleration of zero.
 
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