derive a relation of equation of motion by calculus method

 

Asked by anikettyagi777 | 17th Feb, 2018, 09:21: PM

Expert Answer:

(1) rate of change of velocity = acceleration.
 
Let v be the velocity and a is acceleration. Then mathematically begin mathsize 12px style fraction numerator d v over denominator d t end fraction space equals space a end style
begin mathsize 12px style d v space equals space a space d t integral subscript u superscript v d v space equals space integral subscript 0 superscript t a space d t v space minus space u space equals space a space t space semicolon space w h e r e space u space i s space i n i t i a l space v e l o c i t y space a n d space v space i s space f i n a l space v e l o c i t y h e n c e space v space equals space u plus a space t end style
 
------------------------------------------------------
 
(2) rate of displacement = velocity
 
mathematically begin mathsize 12px style fraction numerator d s over denominator d t end fraction space equals space v semicolon end style
begin mathsize 12px style w h e r e space d s space i s space a space s m a l l space d i s p l a c e m e n t space i n space t i m e space d t  d s space equals space v space d t space equals space open parentheses u space plus space a space t close parentheses space d t space equals space u space d t space plus space a space t space d t  integral subscript 0 superscript S d s space equals space integral subscript 0 superscript t u space d t space plus space integral subscript 0 superscript t space a space t space d t  S space equals space u space t space plus space 1 half a space t squared end style
----------------------------------------------------
 
Let distance travelled is S in time t. Let initial velocity is u and final velocity is v.
 
distance = average velocity × time
begin mathsize 12px style S space equals space fraction numerator v plus u over denominator 2 end fraction cross times t sin c e space v space equals space u plus a t comma space w e space s u b s t i t u t e space t space equals space fraction numerator v minus u over denominator a end fraction space i n space t h e space a b o v e space e q u a t i o n t h e n space S space equals space fraction numerator v plus u over denominator 2 end fraction cross times fraction numerator v minus u over denominator a end fraction space equals space fraction numerator open parentheses v squared space minus space u squared close parentheses over denominator 2 a end fraction h e n c e space v squared space equals space u squared minus 2 space a space S end style

Answered by Thiyagarajan K | 17th Feb, 2018, 09:59: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.