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If I observe a change in direction of velocity, but not in speed: What does the acceleration vector look like?

I am confused! The difference vector between two vectors of equal length A has a different length than A. But acceleration is defined by the time derivative of velocity (which is a vector quantity) and thus by the vector difference - or am I wrong?

Is it an inaccuracy that is no longer relevant when the time interval is very small?

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    $\begingroup$ Why not simply differentiating $x(t)= r \cos \omega t, \; y(t)= r \sin \omega t$ twice and see what happens (together with a drawing). $\endgroup$
    – Hyperon
    Commented Feb 28 at 15:10
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    $\begingroup$ Voting to reopen. Clearly a conceptual question not a "do my homework" question. $\endgroup$
    – gandalf61
    Commented Feb 28 at 15:23
  • $\begingroup$ It was a question that came to my mind while I was drawing the differential triangle. The post refers to the question of whether the size of the interval is decisive $\endgroup$
    – Sylvia
    Commented Feb 28 at 15:34
  • $\begingroup$ Clearly conceptual. $\endgroup$
    – Philip Wood
    Commented Feb 28 at 15:36

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The acceleration vector, any instant of the motion, regardless of whether or not the magnitude of the velocity is altered, is always perpendicular to the instantaneous velocity vector. You can work this out using the concept of a "differential triangle".

For example draw a velocity vector $\vec V$ from some point, then draw the changed velocity vector, i.e. another vector of same length and stemming from the same point but rotated by some small angle from the original. Now, draw a vector from the head of the first, to the head of the original, i.e. $\Delta \vec v$. In the limit that the angle approaches an infinitesimal, the vector $\Delta\vec v$ makes a right angle to the vector $\vec v$, so that $\Delta \vec v$ is perpendicular to $\vec v$. Now it only takes an infinitesimally small amount of time for the velocity vector to change by an infinitesimally small angle, so you have: $$\vec a=\lim\limits_{\Delta t\rightarrow 0}{\Delta\vec v\over \Delta t},$$ which is a vector that is equal to the acceleration and is clearly perpendicular to $\vec v$.

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  • $\begingroup$ thanks a lot! That is (?): Acceleration is defined using inifinitesimally small intervals and a vector that is perpendicular to the instantaneous velocity vector. The confusion about the existence of an acceleration vector having a length when speed is constant disappears if you consider the definition of acceleration that refers to infinitely small intervals (with infinitesimally small vector lengths) $\endgroup$
    – Sylvia
    Commented Feb 28 at 16:14
  • $\begingroup$ The acceleration vector is not, in general, dependent on the magnitude of velocity. Geometrically, one can think of it as related to the curvature of the path on which a particle moves; even if the magnitude of velocity is small, a sharp turn indicates a large acceleration. See, for example, discussions on the "TNB" frame. $\endgroup$
    – Albertus Magnus
    Commented Feb 28 at 16:25
  • $\begingroup$ ok. @Albertus Magnus, I don't understand whether the acceleration vector is perpendicular to the v1 (v at t1), or to v2 (v at t2)... How can you tell? $\endgroup$
    – Sylvia
    Commented Feb 28 at 18:58
  • $\begingroup$ @Sylvia In the limit that $\Delta t$ is small, then the acceleration vector is perpendicular to both of the vectors. $\endgroup$
    – Albertus Magnus
    Commented Feb 28 at 22:53
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    $\begingroup$ thanks a lot for your answers! $\endgroup$
    – Sylvia
    Commented Mar 5 at 9:32
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"But acceleration is defined by the time derivative of velocity (which is a vector quantity) and thus by the vector difference - or am I wrong?"

You are not wrong. If the velocity changes from $\vec {v_1}$ to $\vec{v_2}$ the change in velocity, $\vec{\Delta v}$, is $\vec {v_2}-\vec{v_1}$. Its value can be found by representing $\vec {v_1}$ and $\vec{v_2}$ by arrows in the directions of the velocities and of lengths proportional to the velocities, with their tails at the same point. $\vec{\Delta v}$ is represented in magnitude and direction by the arrow that must be drawn to go from the head of the $\vec {v_1}$ arrow to the head of the $\vec {v_2}$ arrow. Clearly $\vec{\Delta v}$ won't usually have the same magnitude as $\vec {v_1}$ or $\vec {v_2}$, even if these two have the same magnitude as each other.

The mean acceleration as the velocity changes from $\vec {v_1}$ to $\vec {v_2}$ is defined by $$\text{mean acceleration}= \frac{\vec{\Delta v}}{\Delta t}$$

The instantaneous acceleration is the limit of the fraction as we consider a smaller and smaller time interval and correspondingly smaller $\vec{\Delta v}$. We write

$$\text{instantaneous acceleration}=\lim \limits_{\Delta t \to 0} \frac{\vec{\Delta v}}{\Delta t}$$

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