Question

The distance x covered by a body in time t is given by the relation s=a+bt+ct 2 what are the dimensions of a,b and c? Also write the quantities they represent.

Answer 1

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}$

• The distance x covered by a body in time t is given by the relation s=a+bt+ct².

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To Find :- }}}}$

• The dimension of a , b and c .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution :- }}}}$

Given that , the distance is given by the relation ,

$\sf\longrightarrow s = a + bt + ct^2$

• We know the dimension of Distance is [L] . Therefore , according to the Principal of Homogeneity , here a , bt and ct² must be dimensionally same as the dimension of Distance i.e. [L] .

So that ,

$\sf\longrightarrow[ s] =[ a] \\\\\\\sf\longrightarrow [L]= a \\\\\\\sf\longrightarrow \red{ a = [L] }$

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Secondly ,

$\sf\longrightarrow[ s] = [bt ] \\\\\\\sf\longrightarrow [L]= b[T] \\\\\\\sf\longrightarrow b = [L]/ [ T] \\\\\\\sf\longrightarrow \red{ b = [LT^{-1}] }$

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Thirdly ,

$\sf\longrightarrow [s] = [ct^2 ]\\\\\\\sf\longrightarrow [L]= c*[T^2] \\\\\\\sf\longrightarrow c = [L] / [T^2] \\\\\\\sf\longrightarrow \red{ c = [ LT^{-2}]}$

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Therefore ,

$\sf\longrightarrow \textsf{\textbf{ a = [L] }} \\\\\\\sf\longrightarrow \textsf{\textbf{ b = [ LT ^{\textsf{\textbf{-1}}} ] }} \\\\\\\sf\longrightarrow \textsf{\textbf{ c = [LT ^{\textsf{\textbf{-2}}} ]}}$

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• Also the dimensional formula of a is same as distance or displacement .
• The dimensional formula of b same as velocity or speed .
• The dimensional formula of c is same as acceleration .

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Answer 2

Answer:

Explanation:idk