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Suppose an object's velocity is $5 \ \text{m/s}$ at $t = 1$ seconds and $8 \ \text{m/sec}$ at $t = 2$ seconds then the acceleration here is $3 \ \text{m/sec$^2$}$ i.e at $t = 1$ seconds the acceleration is $3 \ \text{m/sec$^2$}$. This isn't instantaneous acceleration, right? It is just an acceleration over the interval from 1-2.

Now, instantaneous acceleration means the change in velocity is happening at that instant, say $v_1$, $v_2$ occur at that particular instant (I know we need $t_1$ and $t_2$ and they keep getting infinitely closer).

Suppose at $t = 1$ seconds the velocity is $15 \ \text{m/s}$ (i.e $v(1 \ \text{s}) = 15 \ \text{m/s}$) and the acceleration is $a = 10 \ \text{m/s$^2$}$ (i.e $a(1 \ \text{s}) = 10 \ \text{m/s$^2$}$). Here the acceleration $10 \text{m/s$^2$}$ happened at an instant i.e $v_1$, $v_2$ we assume happened at an instant, because thats what instantaneous means, and that the change doesn't happen over the interval. i.e it doesnt affect other points of time (say $t = 2$ seconds).

  1. So am i right here? and

  2. If the acceleration at every instant(i.e instantaneous acceleration at every instant being same)is constant. How will it affect the other points of time?? how is the change happening here at each instant?

I am looking for practical explanation, do not explain using kinematical equations, please explain with example. Please answer if I'm right or wrong and also the second point.

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    $\begingroup$ Do you know differential calculus? $\endgroup$ – David White Jan 26 at 0:10
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"Instantaneous acceleration" does not mean "acceleration that happens in an instant". It just means "the value of the acceleration at a specific point in time". If you have constant acceleration over an interval like in your first example, then the instantaneous acceleration is the same value ($3 m/s^2$) for each point in that interval.

You may be confusing the term with an instantaneous change of velocity, which is something else (and not possible in reality except as an approximation, involving an impulse).

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  • $\begingroup$ i dont get it,with what you said that in my first example instantaneous acceleration is same value 3m/s^2 for each point. its like you are calling average acceleration as instantaneous acceleration. that example i mentioned had huge time gap of 1 sec. i.e 1-2 secs,whereas the definition of instantaneous acceleration says delta "t" tends to zero. $\endgroup$ – rahul amare Jan 26 at 10:13
  • $\begingroup$ In this case, yes, the instantaneous acceleration happens to be the same as the average acceleration. If the acceleration wasn't constant over the interval, the instantaneous acceleration would be different at each point (bute there would still be only one average). $\endgroup$ – Kristoffer Sjöö Jan 26 at 16:00
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Ok,that is a pretty nice question, but I vary with your thought over the concept.

Ok, when the velocity changes(causing acceleration), it changes by a small measure at every instant. Now that can be indeterminable because every time interval will include change in the smallest unit of time. (Re-read your own question's second paragraph.

So this means that instantaneous acceleration will occur.

As of the second point, time period can affect acceleration and velocity, but they can not change time!


Hope this solves your query.

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Suppose a particle traces out the curve x(t)=a(t)i+b(y)j+c(t)k in Euclidean space, where i,j,k are the three unit vectors, and x is a vector.

The instantaneous velocity is defined as

(Limit(delta_t->0) (x(t+delta_t)-x(t))/(delta_t))evaluated at (t=t_inst)

(both numerator and denominator approaches a small quantity, so the limit does not necessarily diverge).

This is just dx(t)/dt evaluated at the instance of concern. d^x/dt^2 evaluated at instance of concern gives you the instantaneous acceleration.

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    $\begingroup$ Answers, like questions, are expected to use MathJax for all math. Please edit your answer. $\endgroup$ – G. Smith Jan 25 at 22:12
  • $\begingroup$ What @G.Smith said. And there are more helpful MathJax / Latex links on the page I linked you to the other day, in this comment. BTW, you have a typo in your last expression. $\endgroup$ – PM 2Ring Jan 26 at 0:51

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