Question

Force on a particle in one-dimensional motion is given by $F=A v+B t+\frac{C x}{A t+D}$, where $F$ is force, $v$ is speed, $t$ is time, $x$ is position and $A, B, C$ and $D$ are constants. Dimensions of $C$ are $\left[M^{p} L^{q} T^r\right]$. Find $(p-q-r)$.

Answer 1

$\text{dim}(A)=\text{dim}\left(\frac{F}{v}\right) ; \text{dim}(C)=\text{dim}\left[\frac{F(A t+D)}{x}\right]$
$\Rightarrow \text{dim}(C)=\text{dim}\left[\frac{F}{x}\left(\frac{F t}{v}\right)\right]=\text{dim}\left(\frac{F^{2} t}{x v}\right)$
$=\text{dim}\left(\frac{F^{2}}{v^{2}}\right)=\left[M^{2} T^{-2}\right]$