Question

A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total trip, what are the (a) magnitude and (b) direction of the plane's displacement, the (c) magnitude and (d) direction of its average velocity, and (e) its average speed? Give your angles as positive or negative values of magnitude less than 180 degrees, measured from the +x direction (east)

Answer 1

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is $64.59^{\circ}$ South of East

(c) Average velocity magnitude is 570.88 km

(d) The direction of average velocity is $64.59^{\circ}$ South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, $\vec{AB} = 446 km$

Distance moved from B to C in South direction, $\vec{BC} = - 939 km$

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:

$AC = \sqrt{(AB)^{2} + (BC)^{2}}$

$AC = \sqrt{(446)^{2} + (- 939)^{2}}$ = 1039 km

(b) Direction of the displacement is given by:

$tan\theta = \frac{\vec{BC}}{\vec{AB}}$

$\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}$

$64.59^{\circ}$ South of East

(c) Magnitude of the average speed is given by:

$v_{avg} = \frac{AC}{t + t'}$

$v_{avg} = \frac{1039}{1.82} = 570.88 km/h$

(d) The direction of the average velocity is the same as that of the displacement, i.e., $64.59^{\circ}$ South of East.

(e) The average speed of the [plane is given by:

$v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}$

$v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h$