Answer
a)Yes
b)Yes
c)Yes
Work Step by Step
For an equation to be dimensionally consistent, the dimensions of the quantity on the left hand side should be the same as the dimesions of the quantity on the right hand side.
Dimension of x is [L]
Dimension of v is [L$T^{-1}$]
Dimension of t is [T]
Dimension of a is [L$T^{-2}$]
a) x = vt
Left hand side : x has dimension of [L].
Right hand side : vt has dimension [L$T^{-1}$T] = [L].
Therefore this equation is dimensionally correct.
b) x = $\dfrac{1}{2}at^{2}$
Left hand side : x has dimension of [L].
Right hand side :
$at^{2}$ = [L$T^{-2}$ $T^{2}$] = [L].
Numbers are dimensionless.
Therefore this equation is dimensionally correct.
c) t = $(2x/a)^{0.5}$.
Left hand side : t has dimension of [T].
Right hand side :
2x/a = [L]/[L$T^{-2}$] = [$T^{2}$]
$(2x/a)^{0.5}$ = [T]
Therefore this equation is dimensionally correct.
