How using the average velocity gives the exact value of s?

AI Thread Summary
Using average velocity (∆s/∆t) provides an estimate of displacement during a time interval, but it does not yield the exact displacement (∆s) when acceleration is variable. Average velocity represents the mean displacement per unit time rather than the precise distance traveled. In scenarios with constant acceleration, average velocity can closely approximate actual displacement, but discrepancies arise with variable acceleration. Therefore, while average velocity is useful for calculations, it is important to recognize its limitations in providing exact values. Understanding these nuances is crucial for accurate physics problem-solving.
almirza
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Homework Statement



When we got a constant acceleration or a variable acceleration (i.e. when the velocity of the moving object is not constant), we use averge velocity to be equal to (∆s/∆t).
What I was wondering that if for example we got all the velocities so we can calculate the averge of them and the time taken and we need to calculate the displacement moved, how using averge velocity can give us the right answer or the right value for ∆s?
Is it really the exact displacement moved in time ∆t? How?

Homework Equations





The Attempt at a Solution

 
Last edited:
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You average velocity won't give you the exact value for \Delta s. Your average velocity in that formula will give you an average displacement per unit time interval.
 
G01 said:
You average velocity won't give you the exact value for \Delta s. Your average velocity in that formula will give you an average displacement per unit time interval.

Thank you
 

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