The velocity of a particle depends upon the t | vedclass

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(B) Given the equation $v = \sqrt{ab} + bt + \frac{c}{d + t}$.$1$. For the term $bt$: Since $v$ is velocity $([LT^{-1}])$ and $t$ is time $([T])$,the

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distance,acceleration,distance,time

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(B) Given the equation $v = \sqrt{ab} + bt + \frac{c}{d + t}$.$1$. For the term $bt$: Since $v$ is velocity $([LT^{-1}])$ and $t$ is time $([T])$,the dimension of $bt$ must be $[LT^{-1}]$. Thus,$[b][T] = [LT^{-1}] \Rightarrow [b] = [LT^{-2}]$,which represents acceleration.$2$. For the term $\sqrt{ab}$: The dimension of $\sqrt{ab}$ must be velocity $([LT^{-1}])$. Squaring both sides,$[ab] = [L^2T^{-2}]$. Since $[b] = [LT^{-2}]$,we have $[a][LT^{-2}] = [L^2T^{-2}] \Rightarrow [a] = [L]$,which represents distance.$3$. For the term $d+t$: Since we can only add quantities of the same dimension,$d$ must have the dimension of time $([T])$.$4$. For the term $\frac{c}{d+t}$: The dimension of this term must be velocity $([LT^{-1}])$. Since $[d+t] = [T]$,we have $\frac{[c]}{[T]} = [LT^{-1}] \Rightarrow [c] = [L]$,which represents distance.Therefore,the order is $a$ (distance),$b$ (acceleration),$c$ (distance),$d$ (time).

The velocity of a particle depends upon the time $t$ according to the equation $v = \sqrt{ab} + bt + \frac{c}{d + t}$. The physical quantities which are represented by $a, b, c$ and $d$,are in the following order:

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