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?)

asked by David

7 years ago

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1 answer

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.

answered by Henry

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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 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?)

1 answer

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.

Ask a New Question or answer this question.

Henry · Answer

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.