Question

A small plane flies 40.0 km in a direction 60° north of east and then flies 30.0 km in a direction 15° north of east. Use the analytical method to find the total distance the plane covers from the starting point, and the geographic direction of its displacement vector. What is its displacement vector?

Answer 1

Answer:

d= 64.7 km

$\theta = 40.9^{o}$

displacement vector$=r_xi + r_yj = 48.9i + 42.4j$

Explanation:

total distance = 40 + 30 = 70 km

during 1st flight

$r_1 x = 40*cos60$

$r_1 x = 20 km$

$r_1 y = 40*sin60$

$r_1 y = 34.64 km$

during 2nd flight

$r_2 x = 30*cos15$

$r_2 x = 28.9 km$

$r_2 y = 30*sin15$

$r_2 y = 7.76 km$

the two component of r are:

$r_x = r_1x + r_2x = 20 + 28.9 = 48.9 km$

$r_y = r_1y + r_2y = 34.64 + 7.76 = 42.4 km$

Geographical Direction $\theta = tan^{-1}\frac{r_y}{r_x} [tex]\theta = 40.9^{o}$

Displacement d$= \sqrt{r_x^2 + r_y^2}$

                     $d = \sqrt{48.9^2+42.4^2} = 64.7 km$

d= 64.7 km

displacement vector$=r_xi + r_yj = 48.9i + 42.4j$