Bob has a helicopter from the launch pad he flies the following path. First he travels from the launch pad a distance of 24 kilometers at heading of 60 degrees East of South. Then he flies 47 degrees heading 77 degrees West of North. After this he flies 34 kilometers heading 75 degrees West of North. Now he is 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 number degrees north of east - your answer may be greater than 90 degrees)?

Question

Bob has a helicopter from the launch pad he flies the following path. First he travels from the launch pad a distance of 24 kilometers at heading of 60 degrees East of South. Then he flies 47 degrees heading 77 degrees West of North. After this he flies 34 kilometers heading 75 degrees West of North. Now he is 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 number degrees north of east - your answer may be greater than 90 degrees)?

Answer 1

All angles are measured CCW from +x-axis.

Disp. = 24km[330o] + 47[167] + 34[165].
Disp. = (20.78-12i) + (-45.8+10.57i) + (-32.84+8.80i).
Disp. = -57.86 + 7.37i = 58.33km [-7.26o] = 58.33km[172.74o].

Answer 2

To find the displacement vector that Bob needs to take to return directly to the launch pad, we need to break down the path he has taken into two components: the North-South component and the East-West component. We can then add these components together to get the overall displacement vector.

First, let's analyze the North-South component of Bob's path:

- Bob travels 24 kilometers at a heading of 60 degrees East of South. This means he is moving more towards the South than the East. We can find the Southward component by multiplying the distance traveled by the sine of the angle between his heading and the South direction.
- Southward component = 24 km * sin(60) = 24 km * 0.866 = 20.78 km

Now, let's analyze the East-West component of Bob's path:

- After the first leg of his journey, Bob turns 47 degrees heading 77 degrees West of North. This means he is moving more towards the West than the North. To find the Westward component, we multiply the distance traveled by the cosine of the angle between his heading and the North direction.
- Westward component = 47 km * cos(77) = 47 km * 0.230 = 10.81 km

Bob's second leg of the journey doesn't contribute to the North-South or East-West components since it is heading 75 degrees West of North. Therefore, we can exclude it for calculating the displacement vector.

To find the displacement vector, we simply add the North-South and East-West components together:

Displacement vector = (East/West component, North/South component)
Displacement vector = (10.81 km, -20.78 km)

Therefore, Bob needs to take a displacement vector of (10.81 km, -20.78 km) to return directly to the launch pad from his present location.