$\textbf{Question:}$ An automobile accelerates from rest at $1+3\sqrt{t}$ mph/s for $9$ s.
What is its velocity after $9$ seconds?
My Answer Would Be the Following:
$a(t)=1+3t^{1/2}\cdot \frac{mi}{hr\cdot s}\cdot \frac{60 s}{1 min}\frac{60 min}{1 hr}=(1+3t^{1/2})\cdot 60^2 \frac{mi}{hr^2}$
$\implies v=(t+2t^{3/2})\cdot 60^2+c$ where $t$ is time in hours and $v$ is mph.
$\implies v=(t+2t^{3/2})\cdot 60^2$ again where $t$ is time in hours and $v$ is mph.
So, $v=(\frac{u}{60^2}+2(\frac{u}{60^2})^{3/2})\cdot 60^2$ where $u$ is time in seconds and $v$ is mph.
So, \begin{align*} v&=(\frac{9}{60^2}+2(\frac{9}{60^2})^{3/2})\cdot 60^2\\ &=(\frac{9}{60^2}+2(\frac{3}{60})^{3})\cdot 60^2\\ &=(\frac{9}{60^2}+2(\frac{27}{60^3}))\cdot 60^2\\ &=9+\frac{2\cdot 27}{60}\\ &=9+\frac{54}{60}\\ &= 9.9 \text{ mph}.\\ \end{align*}
The answer in my book is $63$ mph. So, what is wrong with this logic/what is the error?
Notice that the acceleration function itself does not have a meaningful unit. We assume that $t$ is measured in seconds, but certainly $1$ has different units from $\sqrt t$ (in $\text{sec}^{1/2}$). So it does not make sense to do mechanical conversion from one unit system to another. When you started and gave the formula in terms of $\text{mi/hr}^2$, notice that $t$ is still in seconds, not in hours. But you cavalierly switched to $t$ in hours with no attention to the $\sqrt t$.
The problem should have specified that $t$ is measured in seconds, and you were intended just to integrate $a(t)$ in seconds, with the answer then being in mph.