Answer
p = 1
Work Step by Step
Velocity is related to acceleration and distance as
$v^{2}$ = 2a$x^{p}$.
Dimension of distance is [L].
Dimension of velocity is [L$T^{-1}$].
Dimension of acceleration is [L$T^{-2}$].
If the equation is dimensionally consistent, the dimensions of both sides must be the same.
The dimesions of $v^{2}$ are:
$[LT^{-1}]^{2}$ = [$L^{2} T^{-2}$].
The dimensions of a$x^{p}$ are
= [L$T^{-2}$ $L^{p}$] = [$L^{p+1} T^{-2}$].
Numbers are dimensionless.
Now,
[$L^{2} T^{-2}$] = [$L^{p+1} T^{-2}$]
2 = p+1
p= 1.
