Why NaBH4 does not reduce carboxylic acid while LiAlH4 does?
The answer is polarity.

Consider the two ions $\ce{AlH4-}$ and $\ce{BH4-}$ (as they exist as ions in a solution).

Now the electro-negativity of $\ce{Al}$ is $1.5$, and that of $\ce{B}$ is $2$, and that of $\ce{H}$ is $2.1$. 

So the electro-negativity difference ($\Delta\mathrm{EN}$) between $\ce{Al-H}$ bond is $0.6$ and that of $\ce{B-H}$ bond is $0.1$. It is clearly visible that the $\Delta\mathrm{EN}$ of the $\ce{Al-H}$ bond is greater than that of the $\ce{B-H}$ bond.

Now considering the mechanism, due to the small polarity induced by the carbonyl group of the carboxylic acid, the $\ce{H}$ atom in $\ce{Al-H}$ bond is readily broken due to its high polarity, while the $\ce{H}$ in $\ce{B-H}$ can't do the same because it has a low polarity. 

This is the reason why $\ce{NaBH4}$ cannot be used instead of $\ce{LiAlH4}$ for reducing carboxylic acids to alcohols.