Question

$v \propto {a^x}{s^y}$ shows the relation between velocity, acceleration and displacement where $x$ and $y$ equals?
A. $1,\;1$
B. $2,\;1/2$
C. $2,\;1$
D. $1/2,\;1/2$

Answer 1

Hint: We will use dimensional analysis to find the dimensions on both sides of the proportionality. After finding the dimension on the right side of the proportionality, we will equate its exponents with the exponents of the same unit on the left side. Thus we shall obtain two equations of $x$ and $y$. Solving these simultaneous linear equations will give us the required answer.

Complete step by step answer:
Here we will use the dimensional analysis of the given formula to obtain the values of $x$ and $y$. The dimension of velocity $v$ in S.I. units is $\dfrac{{{\text{meter}}}}{{{\text{second}}}}$.
Similarly the dimension of acceleration $a$ in S.I. units is $\dfrac{{{\text{meter}}}}{{{\text{secon}}{{\text{d}}^{\text{2}}}}}$.
In the question above, the equation $v \propto {a^x}{s^y}$ has an exponent of $x$ upon the acceleration. Therefore the dimension of acceleration ${a^x}$ in S.I. units is $\dfrac{{{\text{mete}}{{\text{r}}^{\text{x}}}}}{{{\text{secon}}{{\text{d}}^{{\text{2x}}}}}}$.
Similarly the dimension of displacement $s$ is $\;meter$ in S.I. system of units. However in the question above, the equation $v \propto {a^x}{s^y}$ has an exponent of $y$ upon the displacement. Therefore the dimension of displacement ${s^y}$ in S.I. units is ${\text{mete}}{{\text{r}}^{\text{y}}}$.
The net dimension of ${a^x}{s^y}$ therefore equals $\dfrac{{{\text{mete}}{{\text{r}}^{\text{x}}}}}{{{\text{secon}}{{\text{d}}^{{\text{2x}}}}}} \times {\text{mete}}{{\text{r}}^{\text{y}}}$.
On simplifying we get the dimension as $\dfrac{{{\text{mete}}{{\text{r}}^{{\text{x + y}}}}}}{{{\text{secon}}{{\text{d}}^{{\text{2x}}}}}}$.
This should be equal to the dimension of the velocity which is $\dfrac{{{\text{meter}}}}{{{\text{second}}}}$.
Now equating the exponents of ${\text{meter}}$on both the sides of the proportionality, we have
$x + y = 1$.
Similarly, equating the exponents of ${\text{second}}$ on both sides of the proportionality, we get
$2x = 1$.
Thus we have the two equations as,
$x + y = 1$ and
$2x = 1$.
From the second equation we get
$x = \dfrac{1}{2}$.
Substituting this value in the first equation, we get
$\dfrac{1}{2} + y = 1$, or
$y = \dfrac{1}{2}$.
Thus the values of $x$ and $y$ as obtained are $x = \dfrac{1}{2}$ and $y = \dfrac{1}{2}$.
Therefore, the correct answer is option (D).

Note: Here we have used dimensional analysis to find the correct values of $x$ and $y$. Here we have to assume that the proportionality equation does not include any other such variables, but only includes a proportionality constant.