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Expert-verified Found in: Page 16 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position as ${\mathbit{x}}{\mathbf{=}}{\mathbit{k}}{{\mathbit{a}}}^{{\mathbf{m}}}{{\mathbit{t}}}^{{\mathbf{n}}}$, where ${\mathbit{k}}$ is a dimensionless constant. Show by dimensional analysis that this expression is satisfied if ${\mathbit{m}}{\mathbf{=}}{\mathbf{1}}$ and ${\mathbit{n}}{\mathbf{=}}{\mathbf{2}}$. Can this analysis give the value of ${\mathbit{k}}$?

The expression for the position of the particle satisfies the values of $m=1$,$n=2,$ and $k$ the value cannot be determined.

See the step by step solution

## Step 1: Explanation of Dimensional analysis for a function of particle

The analysis of a relationship between different physical quantities by using the units of measurements and dimensions is called dimensional analysis. It is used to examine the correctness of an equation.

Dimensional analysis is used for verifying the correctness of anexpression and in changing the units from one system to another.

## Step 2: Expression and Dimensions for the position of a particle

Write the expression for the position of the particle as a function of time and acceleration.

$x=k{a}^{m}t{}^{n}\dots \dots \text{}\left(1\right)$

Here, $x$ is the position of the particle, $a$ is the acceleration, $m$ and $n$ are the integer values, $k$ is the constant, and $t$ is the time.

Write the individual dimensions for each physical quantity in the above equation.

The dimension of the position of the particle $x$ is $L$.

Here, $L$ is the dimension of length.

The dimension of the acceleration of the particle $a$ is ${\text{LT}}^{\text{- 2}}$.

Here, T is the dimension of time.

The dimension of the time of the particle $t$ is T. Since $k$ is the constant, it has no dimensions.

## Step 3: Verification of correctness of equation (1) by using the m and n values

Replace for $x$, ${\text{LT}}^{\text{- 2}}$ for $a$, and T for $t$ in equation (1).

$\text{L =}{\left({\text{LT}}^{\text{- 2}}\right)}^{\text{m}}{\left(\text{T}\right)}^{\text{n}}\phantom{\rule{0ex}{0ex}}{\text{= L}}^{\text{m}}{\text{T}}^{\text{- 2m}}{\text{T}}^{\text{n}}\phantom{\rule{0ex}{0ex}}{\text{= L}}^{\text{m}}{\text{T}}^{\text{n - 2m}}\text{.}$

Replace ${\text{L}}^{\text{1}}{\text{T}}^{\text{0}}$ for L in the left hand side of the above equation, which does not change the term L .

${\text{L}}^{\text{1}}{\text{T}}^{\text{0}}{\text{= L}}^{\text{m}}{\text{T}}^{\text{n - 2m}}$

By equating the powers of the dimension L on both the sides of the above equation, we get:

localid="1663823460549" $\begin{array}{rcl}& & {\text{L}}^{\text{1}}{\text{= L}}^{\text{m}}\\ & & \text{1 = m}\\ & & \end{array}$

Similarly, by equating the powers of the dimension T on both the sides of the equation, we get:

localid="1663823786360" ${\text{T}}^{\text{0}}{\text{= T}}^{\text{n - 2m}}\phantom{\rule{0ex}{0ex}}0=n-2m\phantom{\rule{0ex}{0ex}}2m=n$

Since $m=1$, replace the value of $m$ above to find the $n$ value.

$2·1=n\phantom{\rule{0ex}{0ex}}2=n$

From the above calculation of $m=1$ and $n=2,$ the expression for the position of the particle is satisfied using the dimensions of the physical quantities.

Hence, the expression $x=k{a}^{m}{t}^{n}$ satisfies for the $m=1$ and $n=2$ values using the dimensional analysis, and since is a dimensionless constant, its value cannot be determined. ### Want to see more solutions like these? 