Answer:
a.1080km
b.63.4° S of E
c.480km/h
d.63.4° S of E
e.644km/h
Explanation:
a. Using Pythagorean theorem:
![|\overrightarrow AC| =\sqrt{483^2+(-966)^2} =1080.02\\\\\approx1080km](https://tex.z-dn.net/?f=%7C%5Coverrightarrow%20AC%7C%20%3D%5Csqrt%7B483%5E2%2B%28-966%29%5E2%7D%20%3D1080.02%5C%5C%5C%5C%5Capprox1080km)
The planes magnitude is therefore 1080km
b. Given the plane's directions as 483km east and 966 west, the angle of displacement can be obtained using Trigonometric properties as:
![\theta= tan({\frac{-966}{483}})^-^1\\\approx -63.4\textdegree](https://tex.z-dn.net/?f=%5Ctheta%3D%20tan%28%7B%5Cfrac%7B-966%7D%7B483%7D%7D%29%5E-%5E1%5C%5C%5Capprox%20-63.4%5Ctextdegree)
The plane's displacement is therefore given as 63.4° S of E
c. From a, above we know the plane's magnitude as 1080km.
#We also know the time taken as 1.5h. By definition average velocity is distance over time:
#D-displacement, t=0.75+1.5=2.25h
![v_a_v_g=\frac{1080}{1.5^2}\\\\=480km/h](https://tex.z-dn.net/?f=v_a_v_g%3D%5Cfrac%7B1080%7D%7B1.5%5E2%7D%5C%5C%5C%5C%3D480km%2Fh)
Hence, the average velocity of the plane is 480km/h
d. Given the plane's directions as 483km east and 966 west, the angle of displacement can be obtained using Trigonometric properties as:
![\theta= tan({\frac{-966}{483}})^-^1\\\approx 63.4\textdegree](https://tex.z-dn.net/?f=%5Ctheta%3D%20tan%28%7B%5Cfrac%7B-966%7D%7B483%7D%7D%29%5E-%5E1%5C%5C%5Capprox%2063.4%5Ctextdegree)
The plane's displacement is therefore given as 63.4° S of E
e. Average speed is the total distance divided by total time taken.
Total time =45min+1.5h=2.25
Total distance=483+966=1449km
Therefore the speed is calculated as:
![speed=\frac{d_t_o_t_a_l}{t_t_o_t_a_l}\\=1449/2.25\\=644km/h](https://tex.z-dn.net/?f=speed%3D%5Cfrac%7Bd_t_o_t_a_l%7D%7Bt_t_o_t_a_l%7D%5C%5C%3D1449%2F2.25%5C%5C%3D644km%2Fh)
The average speed of the plane is 644km/h