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Q13P

Expert-verifiedFound in: Page 16

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 **${\mathit{x}}{\mathbf{=}}{\mathit{k}}{{\mathit{a}}}^{{\mathbf{m}}}{{\mathit{t}}}^{{\mathbf{n}}}$**, where ${\mathit{k}}$** **is a dimensionless constant. Show by dimensional analysis that this expression is satisfied if ${\mathit{m}}{\mathbf{=}}{\mathbf{1}}$** **and** ${\mathit{n}}{\mathbf{=}}{\mathbf{2}}$**. Can this analysis give the value of ${\mathit{k}}$****?**

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

**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.

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.

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.

$2\xb71=n\phantom{\rule{0ex}{0ex}}2=n$

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.

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