Question

check the correctness of equation using dimensions analysis of s=ut-1/2at²​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\textsf{ \implies Second equation of motion i.e. = \sf s = ut +\dfrac{1}{2}at^2 }$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Check :- }}}$

$\textsf{ \implies The correctness of the equation dimensionally.}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

The equation given to us is s = ut + ½at². This is the second equation of motion which establishes the relationship between acclⁿ , displacement , time and initial velocity . Here , a is acceleration , s is displacement , t is time and u is initial velocity. So ,

$\sf\longrightarrow Dimensional \ formula \ of \ velocity =[v]= \red{ [ M^0 L^1 T^{-1}]}$

$\sf\longrightarrow Dimensional \ formula \ of \ Displacement =[s] =\red{ [ M^0 L^1 T^{0}]}$

$\sf\longrightarrow Dimensional \ formula \ of \ accl^n=[a]=\red{ [ M^0 L^1 T^{-2}]}$

$\sf\longrightarrow Dimensional \ formula \ of \ time=[t]=\red{ [ M^0 L^0 T^{1}]}$

$\underline{\purple{\textsf{Hence here we have in LHS as , }}}$

$\sf :\implies\boxed{ \pink{ \sf [s] = [ M^0L^1T^0]}}$

$\rule{200}2$

$\underline{\purple{\textsf{Hence here we have in RHS as , }}}$

$\sf:\implies [u][t]+ \dfrac{1}{2}[a][t^2] =( [ M^0 L^1T^{-1}])([ M^0L^0T^1]) +( [ M^0 L^1 T^{-2} ])( [ M^0L^0T^1]^2) \\\\\sf:\implies [u][t]+ \dfrac{1}{2}[a][t^2] = [ M^0 L^1 T^0 ] +( [ M^0 L^1 T^{-2}] )( M^0 L^0 T^2]) \\\\\sf:\implies [u][t]+ \dfrac{1}{2}[a][t^2] = [ M^0 L^1 T^0 ] + [ M^{(1+0)} L^{(1+0)} T^{(-2+2)}] \\\\\sf:\implies \boxed{\pink{\sf [u][t]+ \dfrac{1}{2}[a][t^2] = [ M^0 L^1 T^0 ] + [ M^0 L^1 T^0 ] }}$

$\textsf{ \to Since the dimensions of LHS and RHS is same,}\\\textsf{ i.e.\red{[ \sf M^0L^1T^{0} ]} .Hence the formula is dimensionally correct.}$

$\underline{\underline{\textsf{\textbf{ \blue{Hence \ Proved \ ! }}}}}$