Query Regarding Instantaneous Velocity and Instantaneous Acceleration

Question

Suppose an objects velocity is 5m/s at t=1 sec and 8m/sec at t=2 sec then the acceleration here is 3m/sec^2.i.e at t=1 sec the acceleration is 3m/sec^2.this isnt 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 v1 v2 occur at that particular instant(i know we need t1 and t2 and they keep getting infinitely closer).

Suppose at t=1 sec the velocity is 15m/s[i.e v(1s)=15m/s] and the acceleration is a=10m/s^2 [i.e a(1s)=10m/s^2]. here the acceleration 10m/s^2 happened at an instant i.e v1 v2 we assume happened at an instant, cz thats what instantaneous means,and that the change doesnt happen over the interval. i.e it doesnt affect other points of time(say t=2 sec)

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??



Plz explain with example. plz do read whatever i have written from the begining and then answer if m right or wrong and also the second point. would be really grateful for the help.thank you.

Answer 1

As an example, considering a car that is traveling while sometime speeding up, sometimes slowing down.

So its acceleration is constantly changing.

1) Here is an explanation of "instantaneous acceleration" that is as close to rigorous as I can get without using the terminology of calculus.

You know that average acceleration in the interval t1 to t2 is (v2-v1)/(t2-t1).

Here is how you can understand instantaneous acceleration at some particular fixed time t1.

Calculate average acceleration over the interval of 1 second starting at t1.

Calculate average acceleration over the interval of 1/2 second starting at t1.

Calculate average acceleration over the interval of 1/4 second starting at t1.

Calculate average acceleration over the interval of 1/8 second starting at t1.

Calculate average acceleration over the interval of 1/16 second starting at t1.

And so on.

You will observe that the values of average acceleration are getting closer and closer to each other.

(There is no guarantee of that in mathematics, but it is so in physics.)

You will also observe that there a value a0 to which the average accelerations are getting arbitrarily close.

(By that I mean that no matter how small a tolerance you chose, you get find an average acceleration within that tolerance by choosing a sufficiently small interval).

It is possible that none of the average accelerations actually equal a0, but a0 is defined to be the instantaneous acceleration.

2) Here is another attempt at explaining it.

Imagine that somebody asks the following question.

"Over a time interval, I can calculate the average distance the car has travelled,

and I can also calculate its average speed.

But also, at every point in time the car has travelled some instantaneous distance, and had some instantaneous speed.

What does 'instantaneous' mean for distance or speed?"

If you can answer his question for distance or for speed then you can also answer your question for acceleration.

Now let me respond to your statements.

You: "Suppose an objects velocity is 5m/s at t=1 sec and 8m/sec at t=2 sec then the acceleration here is 3m/sec^2.i.e at t=1 sec the acceleration is 3m/sec^2."

Me: No, it does not mean that "at t=1 sec the acceleration is 3m/sec^2".

3m/sec^2 is the average acceleration over the interval from 1 sec to 2 sec.

It is perfectly possible that at time t=1 sec the acceleration was different.

What is true is that during that one second interval there must be a time instant when the instantaneous acceleration happens to be the same as the average 3m/sec^2.

You: "this isnt instantaneous acceleration.right?it is just an acceleration over the interval from 1-2."

Me: Yes.

You: "Now,instantaneous acceleration means the change in velocity is happening at that instant"

Me: Yes.

You: ",say v1 v2 occur at that particular instant"

Me: No, you cannot have two different velocities at any one particular instant.

You: "(i know we need t1 and t2 and they keep getting infinitely closer)."

Me: Yes.

You: "Suppose at t=1 sec the velocity is 15m/s[i.e v(1s)=15m/s] and the acceleration is a=10m/s^2 [i.e a(1s)=10m/s^2]. here the acceleration 10m/s^2 happened at an instant"

Me: Yes.

You: "i.e v1 v2 we assume happened at an instant"

Me: No, you cannot have two different velocities at any one particular instant.

You: ", cz thats what instantaneous means,and that the change doesnt happen over the interval."

Me: Yes, instantaneous means something that does not happen over an interval.

You: "i.e it doesnt affect other points of time(say t=2 sec)"

Me: No, instantaneous acceleration at one point in time does affect velocity at future points in time.

You: "1)So am i right here?"

Me: Sometimes yes, sometimes no.

You: "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??"

Me: If the instantaneous acceleration is the same over an interval then that common acceleration value is also the average acceleration over the interval.

And you know the affect of average acceleration.