Question

Find the dimensions of a, b, c ,d
X=at²+bt +c/d+t where x is distance and t is time​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

$\bold{X = at^{2} + bt + \frac{c}{d + t}}$

$\huge{\bold{\underline{Explanation:-}}}$

By the principle of homogeneity.

${ 1.X = at^{2}}$

${ 2.X = bt}$

${ 3.X = \frac{c}{d + t}}$

• Case-1

${X = at^{2}}$

${ \to L = a T^{2}}$

${ \to \frac{L}{T^{2}} = a}$

${\boxed{\boxed{ a = L T^{-2}}}}$

So, the dimensions if a is ${ L T^{-2}}$

• Case-2

${X = bt}$

${ \to L = b T}$

${ \to b = \frac{L}{T}}$

${\boxed{\boxed{ b = L T^{-1}}}}$

So, the dimensions if b is ${ L T^{-1}}$

• Case-3

By principle of homogeneity

d + t = t.

${d = T}$

${\boxed{\boxed{ d =T}}}$

So, the dimensions if d is ${ T}$

• Case-3

${X = \frac{c}{t}}$

${ \to L = \frac{c}{T}}$

${ \to LT = c}$

${\boxed{\boxed{c = LT}}}$

So, the dimensions if c is ${ L T}$

Answer 2

Answer

To find the dimensions of a,b,c and d from the equation:

$\mathtt{x = at {}^{2} + bt + \frac{c}{d + t} }$

According to Principle of Homogeneity,

• x = at²

$\rightarrow \ \sf{L = a{T}^{2}}\\</p><p>\\ \rightarrow \ \boxed{\sf{a=L{T}^{-2}}}$

• x = bt

$\rightarrow \ \sf{L = bT} \\ \\ \rightarrow \ \boxed{\sf{b = L{T}^{-1}}}$

• x = c/d + t

$\rightarrow \ \sf{L = \frac{c}{T}} \\ \\</p><p>\rightarrow \boxed{\sf{c = LT}}$

• d + t = T

$\rightarrow \ \boxed{\sf{d = T}}$