Answer
a) Yes
b) Yes
c) No
d) Yes
Work Step by Step
Dimension of length is [L].
Dimension of v is [L$T^{-1}$].
Dimension of t is [T].
Dimension of a is [L$T^{-2}$].
a) vt
[L$T^{-1}$][T] = [L].
Thus, vt has dimension of length.
b) $\dfrac{1}{2}at^{2}$
$at^{2}$ = [L$T^{-2}$ $T^{2}$] = [L].
Numbers are dimensionless,
thus, $\dfrac{1}{2}at^{2}$ has dimension of length.
c)2at
[L$T^{-2}$][T] = [L$T^{-1}$] is not the same as the dimension of length,
Numbers are dimensionless,
thus, 2at does not have dimension of length.
d)$\dfrac{v^{2}}{a}$
$\dfrac{[LT^{-1}]^{2}}{[LT^{-2}]}$ = $\dfrac{[L^{2}T^{-2}]}{[LT^{-2}]}$ = [L].
Thus,$\dfrac{v^{2}}{a}$ has dimension of length.
