The instantaneous acceleration $\textbf{a}(t)$ of a particle is defined as the rate of change of its instantaneous velocity $\textbf{v}(t)$: $$\textbf{a}(t)=\frac{\mathrm{d}}{\mathrm{d}t}\textbf{v}(t).\tag{1}$$ If the speed is constant, then $$\textbf{a}(t)=v\frac{\mathrm{d}}{\mathrm{d}t}\hat{\textbf{n}}(t)\tag{2}$$ where $\hat{\textbf{n}}(t)$ is the instantaneous direction of velocity which changes with time.


  • According to the definition (1) what is a deceleration?

  • In case (2), when will $\textbf{a}(t)$ represent a deceleration? For example, in uniform circular motion, why is it called the centripetal acceleration and not centripetal deceleration?

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    $\begingroup$ The x and y components of velocity can change over time in such a way that the total speed remains constant over time. $\endgroup$
    – user140374
    May 4, 2018 at 11:17
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    $\begingroup$ Of all questions, why is it this one that gets popular?. Why do laymen enjoy reading about semantics so much? Do they not know what a dictionary is? $\endgroup$
    – knzhou
    May 5, 2018 at 8:44
  • $\begingroup$ @knzhou What I mean is that "deceleration" is not a word with a definite technical meaning That itself wasn't known to me. My doubt was that deceleration could have a meaning which can be extracted as a special case from (2). Which one is "official" and which one is not, may not be known to everyone. But I'm glad that you at least left a comment before downvoting. $\endgroup$
    – SRS
    May 5, 2018 at 13:13

4 Answers 4


Acceleration is the general term for a changing velocity. Deceleration is a kind of acceleration in which the magnitude of the velocity is decreasing. The reason this might be confusing is because the word 'acceleration' is sometimes used to mean that the magnitude of the velocity is increasing, to contrast it with deceleration. One cannot go wrong, however, if one always takes acceleration to mean simply 'changing velocity'. In that case, circular motion corresponds to acceleration (because the velocity is changing) but not deceleration (because its magnitude is not decreasing).

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    $\begingroup$ While it is true that people often use "deceleration" to mean decreasing speed this is problematic in that a particular acceleration could be both increasing and decreasing the magnitude of velocity depending on the frame of reference you chose. $\endgroup$ May 4, 2018 at 15:39
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    $\begingroup$ "Deceleration" is linguistics, not physics. $\endgroup$
    – jamesqf
    May 4, 2018 at 17:45
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    $\begingroup$ Your car is rolling backward down a hill towards a raging river. Your brakes have failed. You want the magnitude of your velocity to go to zero. Don't you push on the accelerator? $\endgroup$
    – DJohnM
    May 4, 2018 at 23:27
  • $\begingroup$ @DJohnM Exactly? $\endgroup$
    – somebody
    May 5, 2018 at 2:34
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    $\begingroup$ @DJohnM In that case I just release the clutch slightly to slow down, no need to press the accelerator unless the hill is really steep $\endgroup$
    – Ferrybig
    May 5, 2018 at 16:52

Acceleration is the correct technical term for the physical quantity you mentioned in the equations you posted (i.e. a).

The term deceleration doesn't describe a rigorously-defined standard physical quantity, it's just a term used differently in different situations that means "handwavily" that the velocity or speed is decreasing.

Sometimes it could be clear that it refers to some precise quantity (e.g. the absolute value of a scalar acceleration along a curve, like when you are driving a car and keep an eye on the odometer), but without further context it has no rigorous meaning.


According to this definition, "deceleration" is undefined.

  • $\begingroup$ To the downvoter: Perhaps you can explain why this answer is not responsive to "According to the definition (1) what is a deceleration?" and therefore responsive to the use of this undefined term in OP's (2). $\endgroup$ May 5, 2018 at 19:21
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    $\begingroup$ I didn't downvote, but any answer this short is bound to attract downvotes. Your comment is more detailed than your answer! $\endgroup$ May 27, 2018 at 2:33

Acceleration's the rate of change of a body's velocity.

  • Deceleration refers to the decrease in the absolute value of the velocity. Definition (1) defines a case of deceleration when $a(t)<0$ when $\mathbf v(t)>0$, or $a(t)>0$ when $\mathbf v(t)<0$. It's just a fancy word to describe a body which is slowing down.
  • In definition 2, $v$ will always be positive, because speed is a scalar quantity. But the direction of change of velocity can be negative. Consider a body moving with a constant speed in a curved path on the following plane. enter image description here Even though the vertical component of velocity is positive, it is constantly decreasing, so we can call this deceleration along the y axis, even though we have simultaneous acceleration along the x axis. $\mathbf v_\mathbf y=v\times cos(kt)$, where k is some constant.Putting this in equation 2,$\textbf{a}(t)=-vk \times sin(kt)$, which is negative for the range of $t$ values we're considering.
  • For circular motion, we call it centripetal acceleration because it's always positive. the formula, which is derived here: A simple derivation of the Centripetal Acceleration Formula?, is $a=v^2/r$. $v$ is speed, which is a scalar and is thus always positive, and $r$ is radius, which is positive again, so $a>0$.

However, in physics (from what I've seen), we don't use the word deceleration frequently, because as I showed earlier, it's simple when the velocity is positive, but when we're dealing with a body with a negative and changing velocity, it gets messy. It's more suitable for explaining things we see, and isn't easily compatible with math.

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    $\begingroup$ Negative acceleration isn't necessarily deceleration, if the body is traveling in the negative direction. In the case of throwing a ball vertically, the acceleration due to gravity is constant. For the first half of the trip, that constant acceleration decelerates the ball (decreasing speed), and then accelerates it for the second half (increasing speed). $\endgroup$ May 4, 2018 at 12:33
  • $\begingroup$ Thanks! I'll add a little clearer definition which emphasizes that the absolute value of the velocity is decreasing. $\endgroup$
    – user191954
    May 4, 2018 at 12:36
  • $\begingroup$ As alluded to in the other answer, the choice of rest frame makes a difference. Consider a car going from rest to 90km/h relative to the ground, while a train travels in the same direction at a constant 50km/h relative to the ground. From a ground observer's viewpoint, the absolute value of the car's speed increases monotonically. From a train observer's viewpoint, it decreases from -50 to zero, then continues to increase from zero to +40. Even though the sign of $$dv/dt$$ is always positive, the train observer might call this deceleration then acceleration. $\endgroup$ May 4, 2018 at 17:55

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