Bob has a helicopter and from the launch pad, he flies the following path. First. he travels from the launch pad a distance of 20km at a heading of 30 degrees East of South (240 degrees). Then he flies 45 km heading 65 degrees West of North (155 degrees). After this, he flies 50 km heading 70 degrees West of North (160 degrees). Now he's ready to return to the launch pad. What is the displacement vector that he needs to take to return directly to the launch pad, from his present location (for the heading, give the # of degrees north of east- your answer may be greater than 90 degrees)?

Question

Bob has a helicopter and from the launch pad, he flies the following path. First. he travels from the launch pad a distance of 20km at a heading of 30 degrees East of South (240 degrees). Then he flies 45 km heading 65 degrees West of North (155 degrees). After this, he flies 50 km heading 70 degrees West of North (160 degrees). Now he's ready to return to the launch pad. What is the displacement vector that he needs to take to return directly to the launch pad, from his present location (for the heading, give the # of degrees north of east- your answer may be greater than 90 degrees)?

Answer 1

To solve this problem, we will break down Bob's flight path into its x and y components. We can then add up these components to find the displacement vector that Bob needs to take to return to the launch pad.

Let's begin by visualizing Bob's flight path. From the launch pad, he first flies 20 km at a heading of 30 degrees (East of South), then 45 km at a heading of 65 degrees (West of North), and finally 50 km at a heading of 70 degrees (West of North).

First, let's consider the x component of each leg of the flight path. When Bob flies "East of South," it means he is heading towards the east but at a downward angle from the south. This can be calculated using trigonometry. To find the x component, we use the formula: x = distance * cos(angle).

For the first leg, Bob travels 20 km at a heading of 30 degrees East of South. The x component can be calculated as follows:
x_1 = 20 km * cos(30 degrees).

Next, we calculate the x component of the second leg. Bob travels 45 km at a heading of 65 degrees West of North. The x component can be calculated as follows:
x_2 = 45 km * cos(180 degrees - 65 degrees).

Lastly, we calculate the x component of the third leg. Bob travels 50 km at a heading of 70 degrees West of North. The x component can be calculated as follows:
x_3 = 50 km * cos(180 degrees - 70 degrees).

Now, let's consider the y component of each leg of the flight path. When Bob flies "West of North," it means he is heading towards the west but at an upward angle from the north. This can also be calculated using trigonometry. To find the y component, we use the formula: y = distance * sin(angle).

For the first leg, Bob travels 20 km at a heading of 30 degrees East of South. The y component can be calculated as follows:
y_1 = 20 km * sin(30 degrees).

Next, we calculate the y component of the second leg. Bob travels 45 km at a heading of 65 degrees West of North. The y component can be calculated as follows:
y_2 = 45 km * sin(180 degrees - 65 degrees).

Lastly, we calculate the y component of the third leg. Bob travels 50 km at a heading of 70 degrees West of North. The y component can be calculated as follows:
y_3 = 50 km * sin(180 degrees - 70 degrees).

To find the displacement vector that Bob needs to take to return directly to the launch pad, we sum up the x and y components of each leg of the flight path. The displacement vector can be calculated as follows:
displacement_vector = (x_1 + x_2 + x_3) * i + (y_1 + y_2 + y_3) * j.

Now, you can plug in the values calculated for x and y components to find the displacement vector.