Question

In the formula x =at+bt^2+ct^3, x=distance distance and t=time. Find the dimensions of a,b and c.

Answer 1

Given:

Formula provided as :

x = at + bt² + ct³

x has a dimension of distance , t has a dimension of time .

To find:

Dimensions of a , b , c .

Concept:

For an equation to be dimensionally correct , the LHS and RHS part has to have the same dimensions. Also only quantities having similar dimensions can be added.

Calculation:

$\therefore \: \: \{x\} = \{at\}$

$\implies\: \: \{L\}= \{aT\}$

$\implies\: \: \{a\} = \{L{T}^{-1}\}$

Similarly ,

$\therefore \: \: \{x\} = \{b{t}^{2}\}$

$\implies \: \: \{L\}= \{b{T}^{2}\}$

$\implies \: \: \{b\} = \{L{T}^{-2}\}$

Similarly ,

$\therefore \: \: \{x\} = \{c{t}^{3}\}$

$\implies \: \: \{L\} = \{c{T}^{3}\}$

$\implies \: \: \{c\} = \{L{T}^{-3}\}$

Answer 2

$\bigstar$ Answer:

Given:

$\grren{\implies \sf x = at+bt^2+ct^3}$

Where,

$\sf \implies x = distance$

$\sf \implies t = time$

To Find:

$\blue{\implies \sf Dimensions\; of\; a , b , c}$

Solution:

Principle of Homogeneity:

According to this principle the given physical equation is dimensionally correct if all the physical quantities in the equation are same.

This principle follows physical quantity of same kind and same order can be added subtracted and equated.

Hence,

$\implies \rm [x] = [at]$

$\implies \rm [L] = [aT]$

$\implies \rm [a]= [LT^-1]$

$\orange{\boxed{\rm a = LT^-1}}$

$\implies \rm [x] = [bt^2]$

$\implies \rm [L] = [bT^2]$

$\implies \rm [b] = [LT^-2]$

$\purple{\boxed{\rm b = LT^-2}}$

$\implies \rm [x] = [ct^3]$

$\implies \rm [L] = [cT^3]$

$\implies \rm [c] = [LT^-3]$

$\pink{\boxed{\rm c = LT^-3}}$