Asked by Melissa
In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines:
2.5 km 45° north of west; then
4.70 km 60° south of east; then
5.1 km straight east; then
7.2 km 55° south of west; and finally
2.8 km 10° north of east.
What is his final position relative to the island?
2.5 km 45° north of west; then
4.70 km 60° south of east; then
5.1 km straight east; then
7.2 km 55° south of west; and finally
2.8 km 10° north of east.
What is his final position relative to the island?
Answers
Answered by
Steve
2.5 @ W45N = -1.77i + 1.77j
4.7 @ E60S = 2.35i - 4.07j
do the others likewise.
Then add them all up and convert back to a distance and bearing
4.7 @ E60S = 2.35i - 4.07j
do the others likewise.
Then add them all up and convert back to a distance and bearing
Answered by
Henry
All angles are measured CCW from +x-axis.
Disp. = 2.5[135o]+4.7[300]+5.1+
7.2[235]+2.8[10o].
Disp. = -1.77+1.77i + 2.35-4.07i + 5.1 + 4.13-5.90i + 2.76+0.486i
Disp. = 12.6 - 7.71i = 14.8km[ -31.5o] = 14.8km[31.5o] S. of E.
Disp. = 2.5[135o]+4.7[300]+5.1+
7.2[235]+2.8[10o].
Disp. = -1.77+1.77i + 2.35-4.07i + 5.1 + 4.13-5.90i + 2.76+0.486i
Disp. = 12.6 - 7.71i = 14.8km[ -31.5o] = 14.8km[31.5o] S. of E.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.