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(A) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.$1$. For the term $(t+c)$,the

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$LT^{-2}, L, T$

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(A) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.$1$. For the term $(t+c)$,the dimension of $c$ must be the same as the dimension of time $t$. Therefore,$[c] = [T]$.$2$. For the term $at$,the dimension of $at$ must be equal to the dimension of velocity $v$. Since $[v] = [LT^{-1}]$ and $[t] = [T]$,we have $[a][T] = [LT^{-1}]$,which gives $[a] = [LT^{-2}]$.$3$. For the term $\frac{b}{t+c}$,the dimension of this term must be equal to the dimension of velocity $v$. Since $[t+c] = [T]$ and $[v] = [LT^{-1}]$,we have $\frac{[b]}{[T]} = [LT^{-1}]$,which gives $[b] = [L]$.Thus,the dimensions are $[a] = [LT^{-2}]$,$[b] = [L]$,and $[c] = [T]$.

If the equation for the velocity of a particle at time $t$ is $v = at + \frac{b}{t+c}$,then the dimensions of $a, b, c$ are respectively:

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