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

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SUMMARY

In the discussion, participants concluded that it is not possible to maintain constant speed while experiencing constant acceleration. The equation for constant acceleration in three dimensions, ##\vec v = \vec v_0 + \vec a t##, indicates that the magnitude of velocity changes over time. The only scenario where constant speed can occur is in uniform circular motion, where acceleration is directed towards the center, but the magnitude of acceleration can be zero. The conversation emphasizes the importance of distinguishing between constant acceleration and zero acceleration.

PREREQUISITES
  • Understanding of vector calculus, particularly in three dimensions.
  • Familiarity with the equations of motion, specifically ##\vec v = \vec v_0 + \vec a t##.
  • Knowledge of uniform circular motion and its characteristics.
  • Basic principles of differentiation in physics.
NEXT STEPS
  • Study the principles of uniform circular motion and its mathematical representation.
  • Learn about the implications of constant acceleration in different motion scenarios.
  • Explore vector calculus applications in physics, focusing on motion equations.
  • Investigate the relationship between velocity, acceleration, and trajectory in three-dimensional motion.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in the principles of motion, particularly those studying kinematics and dynamics in three-dimensional space.

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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