Question

The distance covered by a particle in time t is given by $x = a + bt + c{t^2}$. Find the dimensions of a, b and c.

Answer 1

Use principle of homogeneity to solve this problem. The principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. Also, two physical quantities either added or subtracted must have the dimensional formula.
So all the terms on the left have the same dimension as of distance. So in order to determine the value of a, b, c we will equate the terms with distance dimensionally.

Complete step by step answer:
According to question
To be able to solve the problem we must know what dimensions are.
Any physical quantity is written in the form of its fundamental quantities namely
Mass= [M]
Length= [L]
Time-= [T]
Current= [A]
Temp= [k] or [θ]
Amount of substance= [mol]
Luminous Intensity= [Cd]
In order to write a Dimensional formula, we must convert the SI units such as
Acceleration SI unit = $meter/{\sec ^2} = m{s^{ - 2}}$
Where, meter is length and second is time.
So the dimensional formula of acceleration will become $\left[ {{M^0}{L^1}{T^{ - 2}}} \right]$
As there is no kg or any mass in the unit of acceleration so the exponent of $M$ is zero
In the question distance x is given varying with time ‘t’ as
$x = a + bt + c{t^2}$
So according to principle of homogeneity we have
Dimension $[x]$ = Dimension $[a]$
Dimension $[x]$ = Dimension $[bt]$
Dimension $[x]$ = Dimension $[c{t^2}]$
$x$ is distance so dimension will be
Dimension $[x]$ = $[{M^0}{L^1}{T^0}]$
Dimension $[t]$ = $[{M^0}{L^0}{T^1}]$
Dimension $[{t^2}]$ = $[{M^0}{L^0}{T^2}]$
Dimension $[bt]$ = Dimension $[b] \times$ Dimension $[t]$ = Dimension $[b] \times [{M^0}{L^0}{T^1}]$
Dimension $[c{t^2}]$ = Dimension $[c] \times$ Dimension $[{t^2}]$= Dimension $[c] \times [{M^0}{L^0}{T^2}]$
Equating all the dimensions we get
Dimension $[a]$ = Dimension $[x]$

$\therefore$ Dimension $[a]$ = $[{M^0}{L^1}{T^0}]$

Dimension $[b]$ = $\dfrac{{Dimension[x]}}{{Dimension[t]}}$ = $\dfrac{{[{M^0}{L^1}{T^0}]}}{{[{M^0}{L^0}{T^1}]}}$

$\therefore$ Dimension $[b]$ = $\left[ {{M^0}{L^1}{T^{ - 1}}} \right]$

Dimension $[c]$ = $\dfrac{{Dimension[x]}}{{Dimension[{t^2}]}}$ = $\dfrac{{[{M^0}{L^1}{T^0}]}}{{[{M^0}{L^0}{T^2}]}}$

$\therefore$ Dimension $[c]$ = $\left[ {{M^0}{L^1}{T^{ - 2}}} \right]$

Note:
This problem can alternatively be solved by another concept.
Since a, bt & ct are being added so their dimensional formula should be the same. So in order to find the dimension, we will equate all the terms in the left side with each other and then we can easily determine the dimension of a, b and c.
As we know the dimension of a is equal to x distance by the law of homogeneity.
Dimension $[a]$ = Dimension $[x]$

$\therefore$ Dimension $[a]$ = $[{M^0}{L^1}{T^0}]$

For dimension of b
Dimension $[a]$ = Dimension $[b] \times$ Dimension $[t]$
Dimension $[b]$ = $\dfrac{{[{M^0}{L^1}{T^0}]}}{{[{M^0}{L^0}{T^1}]}}$ = $[{M^0}{L^1}{T^{ - 1}}]$

$\therefore$ Dimension $[b]$ = $\left[ {{M^0}{L^1}{T^{ - 1}}} \right]$

Similarly for dimension of c
Dimension $[a]$ = Dimension $[c] \times$ Dimension $[{t^2}]$

Dimension $[c]$ = $\dfrac{{Dimension[a]}}{{Dimension[{t^2}]}}$ = $\dfrac{{[{M^0}{L^1}{T^0}]}}{{[{M^0}{L^0}{T^2}]}}$
$\therefore$ Dimension$[c]$ = $\left[ {{M^0}{L^1}{T^{ - 2}}} \right]$