-->

Motion In Straight Line

Question
CBSEENPH11017214

A particle is moving along a straight line such that its displacement at any instant is given by,

s1 = αt1 + 2βt + γ

where α, β and γ are constants.

Find the acceleration of the particle.

Solution

Given here,
   space space space space space space space space space space space space space space straight s squared equals αt squared plus 2 βt plus straight capital upsilon space space space space space space space space space... left parenthesis 1 right parenthesis 
Differentiating (1) w.r.t. t,

space space space space straight d over dt left parenthesis straight s squared right parenthesis space equals space straight d over dt left parenthesis αt squared plus 2 βt plus straight gamma right parenthesis 
rightwards double arrow    2 xv equals 2 αt plus 2 straight beta space space space space space space space space space space space space space space space space space space space space space space open square brackets because dx over dt equals straight v close square brackets
rightwards double arrow        straight v equals fraction numerator αt plus straight beta over denominator straight s end fraction space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
Differentiating (2) w.r.t. t, 
straight d over dt left parenthesis straight v right parenthesis equals straight d over dt left parenthesis fraction numerator αt plus straight beta over denominator straight s end fraction right parenthesis 
That is, acceleration of the particle is given by, 
     space space straight alpha equals fraction numerator straight s begin display style straight d over dt end style left parenthesis straight alpha plus straight beta right parenthesis minus left parenthesis αt plus straight beta right parenthesis begin display style straight d over dt end style left parenthesis straight s right parenthesis over denominator straight s squared end fraction   

     
space space space space space equals space fraction numerator sα minus left parenthesis at plus straight beta right parenthesis straight v over denominator straight s squared end fraction 


space space space space space space space space space space space space space equals space fraction numerator sα minus left parenthesis αt plus straight beta right parenthesis straight v over denominator straight s squared end fraction 
   
space space space space space space space space equals fraction numerator sα minus left parenthesis αt plus straight beta right parenthesis begin display style fraction numerator αt plus straight beta over denominator straight s end fraction end style over denominator straight s squared end fraction
         
equals fraction numerator αs minus left parenthesis αt plus straight beta right parenthesis squared over denominator straight s cubed end fraction
         equals fraction numerator straight alpha left parenthesis αt squared plus 2 βt plus straight gamma right parenthesis minus left parenthesis straight alpha squared straight t squared plus 2 αβt plus straight beta squared right parenthesis over denominator straight s cubed end fraction
This implies,
      straight alpha equals fraction numerator αγ minus straight beta squared over denominator straight s cubed end fraction, is the acceleration of the particle.