Bob has a helicopter and from the launch pad he flies the following path. First he travels from the launch pad a distance of 52 kilometers at heading 79 degrees South of West. Then he flies 26 kilometers heading 22 degrees north of East. After this he flies 47 kilometers heading 25 degrees South of West. Now he is ready to return to the launch pad. What is the displacement vector that he needs to take to travel directly to the launch pad, from his present location?(for the heading give the number of degrees North of East-your answer may be greater than 90 degrees?)
All angles are measured CCW from +x-axis.
Disp. = 52km[259o] + 26[22] + 47[205].
Disp. = (-9.92-51i)+(24.1+9.74i)+(-42.6-19.9i) = -28.42 - 61.16i = 67.44km[65.1o] N. of E.
To calculate the displacement vector to travel directly to the launch pad, we need to find the vector sum of all the individual displacements.
Let's break down the information provided into components.
1. First displacement: 52 kilometers at a heading of 79 degrees South of West.
To find the components, we need to consider the angle with respect to the positive x-axis. Since it is South of West, the angle becomes 180 - 79 = 101 degrees.
The x-component of this displacement is -52 km * cos(101°) and the y-component is -52 km * sin(101°).
2. Second displacement: 26 kilometers at a heading of 22 degrees North of East.
The x-component of this displacement is 26 km * cos(22°) and the y-component is 26 km * sin(22°).
3. Third displacement: 47 kilometers at a heading of 25 degrees South of West.
Similar to the first displacement, the x-component of this displacement is -47 km * cos(155°) and the y-component is -47 km * sin(155°).
Now, to find the displacement vector, we sum the x-components and the y-components separately:
X-component = (-52 km * cos(101°)) + (26 km * cos(22°)) + (-47 km * cos(155°))
Y-component = (-52 km * sin(101°)) + (26 km * sin(22°)) + (-47 km * sin(155°))
Finally, the displacement vector is the combination of the resultant x-component and y-component.