Average acceleration = inst accleration ?

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SUMMARY

The discussion confirms that the average acceleration of a body cannot be unequal to its instantaneous acceleration at any point within a given time interval, provided there are no discontinuities in the acceleration function. This conclusion is supported by the Mean Value Theorem in calculus, which asserts that there exists at least one instant where the instantaneous acceleration equals the average acceleration over that interval. The mathematical representation of average acceleration is provided as aavg = (a1t1 + a2t2) / (t1 + t2), illustrating the relationship between time and acceleration values.

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Can the average acceleration of a body be not equal to the instantaneous acceleration for at least an instant?


I know the answer to the question above is NO. But I find the answer; using Mean value theorem in calculus.
Let me know if there is any answer based on physics.
Thanks
 
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What is the meaning of ' average' ?
 
Let v1=At for a time t1
then v2=Bt+At_1 for an additional time t2.
Then a1=A and a2=B.
However, the [time-weighted] average acceleration is
aavg= (a1t1+a2t2)/(t1+t2)=(At1+Bt2)/(t1+t2).
 
Last edited:
robphy said:
Let v1=At for a time t1
then v2=Bt+At_1 for an additional time t2.
Then a1=A and a2=B.
However, the [time-weighted] average acceleration is
aavg= (a1t1+a2t2)/(t1+t2)=(At1+Bt2)/(t1+t2).


Sorry, I don't see any relation between your answer and my question.
I meant in a given time interval is it possible for a body to has average acceleration 'a' but never reach it as instantaneous acceleration.
 
II'm not sure what question your asking, but within in a given inertval there will always be an instant when the acceleration is equal to the avergae accelartion within that inertval as long as there are no discontinuities in the accelartion as a function of time within that interval.
 

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