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:
[tex]|\overrightarrow AC| =\sqrt{483^2+(-966)^2} =1080.02\\\\\approx1080km[/tex]
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:
[tex]\theta= tan({\frac{-966}{483}})^-^1\\\approx -63.4\textdegree[/tex]
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:
[tex]v_a_v_g=\frac{|D|}{t_t_o_t_a_l}\\\\\frac{|\overrightarrow AC| }{t_t_o_t_a_l}[/tex] #D-displacement, t=0.75+1.5=2.25h
[tex]v_a_v_g=\frac{1080}{1.5^2}\\\\=480km/h[/tex]
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:
[tex]\theta= tan({\frac{-966}{483}})^-^1\\\approx 63.4\textdegree[/tex]
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:
[tex]speed=\frac{d_t_o_t_a_l}{t_t_o_t_a_l}\\=1449/2.25\\=644km/h[/tex]
The average speed of the plane is 644km/h
