Answer:
(a) Magnitude is 1039 km
(b) Direction of the displacement is [tex]64.59^{\circ}[/tex] South of East
(c) Average velocity magnitude is 570.88 km
(d) The direction of average velocity is [tex]64.59^{\circ}[/tex] South of East
(e) Average speed is 759.34 km/h
Solution:
Distance moved from A to B in East direction, [tex]\vec{AB} = 446 km[/tex]
Distance moved from B to C in South direction, [tex]\vec{BC} = - 939 km[/tex]
Time taken to move from A to B, t = 43.0 min = 0.72 h
Time taken to move from B to C, t' = 1.10 h
Now,
(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:
[tex]AC = \sqrt{(AB)^{2} + (BC)^{2}}[/tex]
[tex]AC = \sqrt{(446)^{2} + (- 939)^{2}}[/tex] = 1039 km
(b) Direction of the displacement is given by:
[tex]tan\theta = \frac{\vec{BC}}{\vec{AB}}[/tex]
[tex]\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}[/tex]
[tex]64.59^{\circ}[/tex] South of East
(c) Magnitude of the average speed is given by:
[tex]v_{avg} = \frac{AC}{t + t'}[/tex]
[tex]v_{avg} = \frac{1039}{1.82} = 570.88 km/h[/tex]
(d) The direction of the average velocity is the same as that of the displacement, i.e., [tex]64.59^{\circ}[/tex] South of East.
(e) The average speed of the [plane is given by:
[tex]v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}[/tex]
[tex]v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h[/tex]
