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dt tends to 0 then ds tends to 0 but how it is possible that ds/dt do not go towards 0/0 but attain finite value v

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706 views asked Aug 21, 2014 in Mechanics by banavarun1997 (140 points)

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The values of differentiation (ratioie..,v=ds/dt) as viewed in graph perspective(displacement vs time) is a super super fine approximation value.ie..,we are not having a math or physical tool to exactly find the velocity at a particular point if a body is accelearting as time progresses.

First question yourself how erraneous your distance calculation is??considering exact physical happening with your naked eye can you find by viewing with your eye how much distance a car accelarating at 10m/sq.sec has travelled in millimicron second and keep pencil on the graph sheet pin pointedly that value.ie,,. we humans are limited to comprehend and express physics exactly.even we don't have a super fine sharped pencil to point a value exactly.

So,what we have done is to approximate calculation as nearer to the original exact value,and also we have to be aware of that no object can change its velocity value without a small 0.00000000..........1(can't comprehend how small it can be)time gap.The proof is if a body is accelerating it cannot have same value in suucessive intervals of time.

So in  taking the ratio(ds/dt)  we are more are less limiting to this time interval!!!!!.Also remember that a larger number of small lines can fit into a  curve (I mean graph).It is the length of this line(depending on the graph shape) that determines the small time value.

We do not make ds--->0(tending to zero) ,only dt--->0(tending to zero) to ease the math calculations!!!!!

After all math stuff in some topics aims to how close we are to physics but not viceversa.

Hope this clarifies,

Thanks@Gopikrishna Mogasati.
answered Aug 31, 2014 by mogasatigopikrishna (200 points)

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