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

AI Thread Summary
In constant acceleration motion, maintaining constant speed is deemed impossible, as velocity's magnitude changes over time. The discussion highlights that uniform circular motion is an exception where speed remains constant while acceleration is present. Clarification is made that zero acceleration is a valid constant, but the focus is on scenarios involving non-zero acceleration. The mathematical proof indicates that constant speed can only occur if acceleration is zero. Overall, the consensus is that constant acceleration inherently leads to varying speed unless acceleration itself is null.
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##?
 
Ibix said:
##\vec a=0##?
Then we have no acceleration.
 
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 .
 
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.
 
  • #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##.
 
  • #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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