Physics Asked by rahul amare on August 14, 2021
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).
So am i right here? and
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.
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.
Answered by WebDevLearner on August 14, 2021
"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).
Answered by Kristoffer Sjöö on August 14, 2021
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.
Answered by justcurious92 on August 14, 2021
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