After 30 min., the Ford Expedition is exactly at
A.
$v_\text{separation} = v_{Fx} + v_T$
$v_\text{separation} = 60 \cos 60^\circ + 30$
$v_\text{separation} = 60 ~ \text{kph}$ ← answer
Another Solution (by Calculus)
$s^2 = (30t)^2 + (30 - 60t)^2 - 2(30t)(30 - 60t)\cos 60^\circ$
$s = \sqrt{900t^2 + (30 - 60t)^2 - 30t(30 - 60t)}$
After 30 minutes, t = 0.5 hr
$\dfrac{ds}{dt} = \dfrac{d}{dt}\Big[ \sqrt{900t^2 + (30 - 60t)^2 - 30t(30 - 60t)} \Big]_{\, t = 0.5}$
$\dfrac{ds}{dt} = 60 ~ \text{kph}$ ← answer
Below is the details of differentiation if you are using a calculator with no derivative function.
