Question

link to chapter 1.1. (problem 1.4.51 - 53)
consider the physical quantities with dimensions: s=l, v=lt^{-1}, a=lt^{-2}, and t=t. determine whether each of the equations below are dimensionally consistent.
(a) v^{2}=2as
(b) s = vt+0.5at^{2}
(c) v = s/t
(d) a = v/t
(e) s=int adt

Explanation:

Step1: Analyze dimensions of left - hand side of (a)

The dimension of $v$ is $LT^{-1}$, so the dimension of $v^{2}$ is $(LT^{-1})^{2}=L^{2}T^{-2}$.

Step2: Analyze dimensions of right - hand side of (a)

The dimension of $a$ is $LT^{-2}$ and of $s$ is $L$. So the dimension of $2as$ is $(LT^{-2})\times L = L^{2}T^{-2}$. Since the dimensions of the left - hand side and right - hand side are the same, the equation $v^{2}=2as$ is dimensionally consistent.

Step3: Analyze dimensions of left - hand side of (b)

The dimension of $s$ is $L$.

Step4: Analyze dimensions of right - hand side of (b)

The dimension of $v$ is $LT^{-1}$ and of $t$ is $T$, so the dimension of $vt$ is $(LT^{-1})\times T = L$. The dimension of $a$ is $LT^{-2}$ and of $t^{2}$ is $T^{2}$, so the dimension of $0.5at^{2}$ is $(LT^{-2})\times T^{2}=L$. The sum of the dimensions of the two terms on the right - hand side is $L + L = L$. So the equation $s = vt+0.5at^{2}$ is dimensionally consistent.

Step5: Analyze dimensions of left - hand side of (c)

The dimension of $v$ is $LT^{-1}$.

Step6: Analyze dimensions of right - hand side of (c)

The dimension of $s$ is $L$ and of $t$ is $T$, so the dimension of $\frac{s}{t}$ is $\frac{L}{T}=LT^{-1}$. So the equation $v=\frac{s}{t}$ is dimensionally consistent.

Step7: Analyze dimensions of left - hand side of (d)

The dimension of $a$ is $LT^{-2}$.

Step8: Analyze dimensions of right - hand side of (d)

The dimension of $v$ is $LT^{-1}$ and of $t$ is $T$, so the dimension of $\frac{v}{t}$ is $\frac{LT^{-1}}{T}=LT^{-2}$. So the equation $a = \frac{v}{t}$ is dimensionally consistent.

Step9: Analyze dimensions of left - hand side of (e)

The dimension of $s$ is $L$.

Step10: Analyze dimensions of right - hand side of (e)

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

Answer:

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