Question

displacement of a body is S= at+bt2+ct3 find dimensions of a, b, c​

Answer 1

$\bf{\underline{\underline{Given :}}}$

$\mathtt{S = at + bt^2 + ct^3 }$

$\mathtt{where,\: S\: \:is \: \:displacement}$

$\bf{\underline{\underline{To \:find :}}}$

$\mathtt{dimension\: \:of\: \: a,\: b,\: c}$

$\bf{\underline{\underline{We\: know\: that }}}$

$\fbox{\mathtt{Displacement\: (S) = M^{0}L^{1}T^{0} }}$

All the terms should have the same dimension

$\mathtt{S = at}$

$\mathtt{\rightarrow L = a T}$

$\mathtt{\rightarrow a = \Large{\frac{L}{T}}}$

$\mathtt{\red{\rightarrow a = LT^{-1}}}$

$\mathtt{S = bt^{2}}$

$\mathtt{\rightarrow L = b T^{2}}$

$\mathtt{\rightarrow b = \Large{\frac{L}{T^{2}}}}$

$\mathtt{\green{\rightarrow b = LT^{-2}}}$

$\mathtt{S = ct^{3}}$

$\mathtt{\rightarrow L = c T^{3}}$

$\mathtt{\rightarrow c = \Large{\frac{L}{T^{3}}}}$

$\mathtt{\purple{\rightarrow c = LT^{-3}}}$

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Important dimensional formulas

$\mathsf{\bigstar\: Distance (s) = Length\: (L) }$

$\mathsf{\bigstar\: Displacement = Length\: (L) }$

$\mathsf{\bigstar\: Time (t) = Time\: (T) }$

$\mathsf{\bigstar\: Weight = Mass\: (M) }$

$\mathsf{\bigstar\: Speed = \frac{Distance}{time} </p><p>=\: \frac{L}{T} \:\:or\: \:LT^{-1}}$

$\mathsf{\bigstar\: Velocity (v) = \frac{Displacement}{time} = \: \frac{L}{T} \:\: or\: \:LT^{-1}}$