Question

1. Amazon is experimenting with "drone delivery" and you are asked to determine a drone’s delivery route. The drone will leave the Amazon warehouse, make two deliveries, and then return to the warehouse. The first delivery is 7.10 miles from the warehouse in a direction 25.0O south of west. The second delivery is 11.2 miles, in a direction 38.0O south of east, from the location of the first delivery. After making its second delivery, how far and in what direction must the drone fly to return to the warehouse?

Answer 1

Answer:

Step-by-step explanation:

From the information given:

Let assume that the drone first delivery went to a certain place U, 7.10miles from the warehouse in the S 25.0° W direction.

However, the drone thus proceeds to the second delivery place V which is 11.2 miles in the S 38.0° W direction.

Then, location U can be determined from a graphical point of view as follows:

In the negative x-direction from the warehouse

$U_x = 7.1 \times cos 25^0$

$U_x = 7.1 \times 0.9063$

$U_x = 6.44 \ miles$  

In the negative y-direction from the warehouse

$U_y =7.1 \times sin 25 ^0$

$U_y =7.1 \times 0.4226$

$U_y = 3.00 \ miles$

Also: the position of V with respect to U can be determined as follows:

in the positive x-direction from the warehouse

$V_{UX} = UV \ cos 38$

$V_{UX} = 8.83 \ miles$

In the negative y-direction from the warehouse:

$V_{UY} = UV \ sin 38$

$V_{UY} = 6.89 \ miles$

Similarly, we will need to determine the position of V with respect to the warehouse.

i.e.

In the positive x-direction the warehouse

$V_x = U_x +V_{UX}$

$V_x =$ -6.44 + 8.83 ( since $U_X$ is in the negative direction)

$V_x =$ 2.39 miles

In the positive y-direction the warehouse

$V_Y = U_Y + V_{UY}$

$V_Y = -3.00 + (-6.89)$   ( since both $U_Y \ and \ V_{AY}$ are in the negative direction)

$V_Y =$ -9.89 miles

Therefore, from above, the distance emanating how far is the drone back to the starting point is :

$D = \sqrt{V_X^2+V^2_Y}$

$D = \sqrt{(2.39)^2+(-9.89)^2}$

$D = \sqrt{5.7121+97.8121}$

D = 10.18 miles

The direction of the drone can be deduced by taking the tangent of the trigonometry;

i.e.

$tan \ \theta = {\dfrac{opposite \ direction}{adjacent \ direction}}$

$tan \ \theta = {\dfrac{9.89}{ 2.39}}$

$tan \ \theta =4.138$

$\theta = \tan^{-1}(4.138)$

$\theta = 76.41^0$ in the north_west direction of V