Answer
$v(t)=\frac{1}{2}sec(t)+\frac{1}{2}$
Work Step by Step
Step 1. Given $\frac{dv}{dt}=\frac{1}{2}sec(t) tan(t)$, we can find its general antiderivatives as $v(t)=\frac{1}{2}sec(t) +C$ where $C$ is a constant.
Step 2. Using the initial condition $v(0)=1$, we have $\frac{1}{2}sec(0) +C=1$ which gives $C=1-\frac{1}{2}=\frac{1}{2}$ and we can get the solution as $v(t)=\frac{1}{2}sec(t)+\frac{1}{2}$
