Answer
a). The general direction of the storm's velocity is (2) N-W.
b). Average velocity =$(-21.21x+16.28y) \,mi/hr$
Magnitude of the storm is $R=26.74\, mi/hr$ and $\theta=37.5^{\circ}$ with respect to negative x-direction.
Work Step by Step
At 8.00 pm, the storm was 60 mi N-E of the station,
At 10.00 pm, the storm was 75 mi N of the station.
a). So, the general direction of the storm's velocity is (2) N-W.
b). Average velocity = $\frac{Average\, displacement}{Total \,time\,taken}$
$=\frac{75y-60cos45^{\circ}x-60sin45^{\circ}y}{2 hr}=\frac{-42.43x+32.57y}{2hr}=(-21.21x+16.28y) \,mi/hr$
Therefore, magnitude of the storm is $R=26.74\, mi/hr$ and $\theta=37.5^{\circ}$ with respect to negative x-direction.
