consider the physical quantities with dimensions: s = l, v = lt^{-1}, a = lt^{-2}, and t = t. determine…

Explanation:

Step1: Recall dimension - analysis rules

We substitute the dimensions of each quantity into the equations and check if the dimensions on both sides match.

Step2: Analyze equation (a)

For $v^{2}=2as$, the dimension of $v$ is $LT^{-1}$, so the dimension of $v^{2}$ is $(LT^{-1})^{2}=L^{2}T^{-2}$. The dimension of $a$ is $LT^{-2}$ and the dimension of $s$ is $L$. Then the dimension of $2as$ is $(LT^{-2})\times L = L^{2}T^{-2}$. Since the dimensions of both sides are the same, the equation is dimensionally consistent.

Step3: Analyze equation (b)

For $s = vt^{2}+0.5at^{2}$, the dimension of $v$ is $LT^{-1}$, and the dimension of $t$ is $T$, so the dimension of $vt^{2}$ is $(LT^{-1})\times T^{2}=LT$. The dimension of $a$ is $LT^{-2}$, so the dimension of $at^{2}$ is $(LT^{-2})\times T^{2}=L$. The left - hand side has dimension $L$, but the first term on the right - hand side $vt^{2}$ has dimension $LT$, so the equation is not dimensionally consistent.

Step4: Analyze equation (c)

For $v = s/t$, the dimension of $s$ is $L$ and the dimension of $t$ is $T$. So the dimension of $s/t$ is $L/T = LT^{-1}$, which is the same as the dimension of $v$. The equation is dimensionally consistent.

Step5: Analyze equation (d)

For $a = v/t$, the dimension of $v$ is $LT^{-1}$ and the dimension of $t$ is $T$. So the dimension of $v/t$ is $(LT^{-1})/T=LT^{-2}$, which is the same as the dimension of $a$. The equation is dimensionally consistent.

Step6: Analyze equation (e)

For $s=\int adt$, the dimension of $a$ is $LT^{-2}$ and the dimension of $dt$ is $T$. So the dimension of $\int adt$ is $(LT^{-2})\times T = LT^{-1}$, which is not the same as the dimension of $s$ ($L$). The equation is not dimensionally consistent.

Answer:

(a) Dimensionally consistent (b) Not dimensionally consistent (c) Dimensionally consistent (d) Dimensionally consistent (e) Not dimensionally consistent