Question

check whether the relation
$s = ut + 1 \div 2a {t}^{2}$
is dimensionally correct or not???

Answer 1

hi ✌️

yes, it is dimensionally correct...

let see the solution..☑️☑️☑️


we have ,

$s = ut + 1 \div 2a {t}^{2}$
checking the dimensions on both sides,

L.H.S=
${m}^{0} \: {l}^{1} \: {t}^{0}$
R.H.S=

$l {t}^{ - 1} \: {t} \: + l {t}^{ - 2} \: {t}^{2 } = {m}^{0} \: {l}^{1} \: {t}^{0}$
L.H.S =R.H.S


Hence the formula is dimensionally correct


hope it will help you ☺️