Let $\{X(t), t \geqslant 0\}$ be Brownian motion with drift coefficient $\mu$ and variance parameter $\sigma^{2}$. That is,
$$
X(t)=\sigma B(t)+\mu t
$$
Let $\mu>0$, and for a positive constant $x$ let
$$
\begin{aligned}
T &=\operatorname{Min}\{t: X(t)=x\} \\
&=\operatorname{Min}\left\{t: B(t)=\frac{x-\mu t}{\sigma}\right\}
\end{aligned}
$$
That is, $T$ is the first time the process $\{X(t), t \geqslant 0\}$ hits $x .$ Use the Martingale stopping theorem to show that
$$
E[T]=x / \mu
$$