# Are velocity and acceleration smooth quantities?

My thinking:

1. acceleration corresponds to a force which is instantaneous, so the acceleration of a rigid body can be rather spiky (non-smooth)

2. velocity (angular velocity) describes the ratio of change of the distance(angle), so it is smooth in the real world.

Conclusion, it makes sense to smooth (e.g., simple averaging) a velocity signal (temporal velocity), but it does not make so much sense to do smoothing on acceleration signal. Am I right?

• What is your definition of smooth? – Hritik Narayan Dec 9 '14 at 15:42
• @HritikNarayan thanks for the response and sorry for the ambiguity, I am a physics newbie. I mean "doing smoothing" here: the possibility to use imputation (if at some point the measurement is missing) and smoothing (using techniques for example, simple averaging) – Hello lad Dec 9 '14 at 15:47
• What do you mean by "a force which is instantaneous"? It's true that acceleration at any given instant is determined by the force at the same instant, but in realistic scenarios we do expect the forces on an object to vary in a continuous way, not change instantaneously, so the acceleration should vary continuously too. – Hypnosifl Dec 9 '14 at 16:04
• Possible duplicates: physics.stackexchange.com/q/35674/2451 , physics.stackexchange.com/q/9720/2451 , physics.stackexchange.com/q/1324/2451 and links therein. – Qmechanic Dec 9 '14 at 20:45
• I think this could be a duplicate of the last of those but it doesn't seem like a duplicate of the others. – David Z Dec 10 '14 at 5:22

As far as we know and can test, space is continuous, not discrete. Since space is continuous, then the labels we associate with it (i.e., positions) are also continuous. Calculus requires continuous functions to do the derivative and integral, so this implies that velocities and accelerations are also continuous because they are derivatives of positions: $$v(t)=\frac{dx(t)}{dt}\qquad a(t)=\frac{dv(t)}{dt}=\frac{d^2x(t)}{dx^2}$$