The velocity $v$ of a particle at time $t$ is given by $v=a t^{2}+\frac{b}{t^{2}+c}$, where $a, b$ a

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The velocity $v$ of a particle at time $t$ is given by $v=a t^{2}+\frac{b}{t^{2}+c}$, where $a, b$ and $c$ are constants. The dimensions of $a, b$ and $c$ are

Professor PC Thomas Classes · Answer

(a) : $v=a t^{2}+\frac{b}{t^{2}+c}$ \\ As $c$ is added to $t^{2}, 	herefore \quad[c]=\left[\mathrm{T}^{2}ight]$ \\ $\left[a t^{2}ight]=\left[\mathrm{LT}^{-1}ight] \quad 	ext{ or, } [a]=\frac{\left[\mathrm{LT}^{-1}ight]}{\left[\mathrm{T}^{2}ight]}=\left[\mathrm{LT}^{-3}ight]$ \\ $\frac{[b]}{\left[\mathrm{T}^{2}ight]}=\left[\mathrm{LT}^{-1}ight] \quad 	ext{ or, } [b]=[\mathrm{LT}]$