an airplane is heading due north at 700 kph, and a wind blows at 60 kph in the direction s 45° e. what is…

Answer:

B. 658.94 kph

Explanation:

Step1: Represent velocities as vectors

Let the velocity of the airplane (\vec{v}{a}=700\hat{j}) (north - direction). The wind velocity (\vec{v}{w}): The direction (S45^{\circ}E) means the angle with the negative (y) - axis is (45^{\circ}) towards the positive (x) - axis. The magnitude of the wind velocity (|\vec{v}{w}| = 60) kph. The (x) - component (v{wx}=60\sin45^{\circ}) and the (y) - component (v_{wy}=- 60\cos45^{\circ}) [v_{wx}=60\times\frac{\sqrt{2}}{2}=30\sqrt{2}\approx42.43,\quad v_{wy}=-60\times\frac{\sqrt{2}}{2}=-30\sqrt{2}\approx - 42.43]

Step2: Find the resultant velocity vector

The resultant velocity (\vec{v}=\vec{v}{a}+\vec{v}{w}) (\vec{v}=(30\sqrt{2})\hat{x}+(700 - 30\sqrt{2})\hat{y}) The magnitude of the resultant velocity (|\vec{v}|=\sqrt{(30\sqrt{2})^{2}+(700 - 30\sqrt{2})^{2}}) [=\sqrt{1800+(700^{2}-42000\sqrt{2}+1800)}] [=\sqrt{3600 + 490000-42000\sqrt{2}}] [=\sqrt{493600-42000\times1.4142}] [=\sqrt{493600 - 59396.4}] [=\sqrt{434203.6}\approx658.94] kph