In an attempt to escape a desert island, a castaway builds a...

In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the day, and she is blown along the following straight lines: 3.00 km and 45.0\north of west, then 5.90 km and 60.0\$^\circ$ south of east, then 2.80 km and 25.0\$^\circ$ south of west, then 5.10 km due east, then 1.70 km and 6.00\$^\circ$ east of north, then 7.20 km and 55.0\$^\circ$ south of west, and\finally 2.80 km and 10.0\$^\circ$ north of east. Use the analytical method to find the resultant vector of all her displacement vectors (in m). (Express your answer in vector form. Assume the $+x$-axis is to the\east and the $+y$-axis is to the north.) What is its magnitude (in km) and direction (in degrees south of east)? magnitude km direction $^\circ$ south of east

Transcript

00:01 In this exercise, we have a raft that travels along straight lines.

00:04 First, it travels 2 .5 kilometers with a direction of 45 degrees north of east.

00:13 Then it travels 4 .7 kilometers with a direction of 90, i'm sorry, of 60 degrees south of east.

00:24 Then 1 .3 kilometers in a direction of 25 degrees.

00:30 South of west, then 5 .1 kilometers to the east, then 1 .7 kilometers north, i'm sorry, east of north, then 7 .2 kilometers south of west, and then 2 .8 kilometers north of east.

00:54 Our goal is to find the total magnitude, the total vector, the total displacement vector of the raft, the magnitude and its direction.

01:06 So first of all, what i'm going to do is to set up a coordinate system that has the y -axis pointing to the north, the x -axis to the east, and i'm going to write each one of these vectors in this coordinate system, and then i'm going to sum it all up.

01:27 So what i'm going to do is that i'm going to call the first vector a1, the second one, a2, and then so on.

01:38 So a3, a4, a5, a6, a7.

01:48 Okay.

01:50 So i'm going to expand each vector in its components.

01:55 So a1 is equal to its magnitude 2 .5 kilometers times the cosine of 45.

02:06 Notice that it's in the positive direction, both in the x and in the y axis.

02:12 So it's plus 2 .5 sine of 45 degrees times j.

02:20 A2 is the magnitude.

02:24 Notice that a2 is pointing in the positive.

02:26 Direction in the x direction but in the negative direction in the y -axis.

02:30 So i have 4 .7 times a cosine of 60 i minus 4 .7 times the sign of 60 j.

02:45 Then we have a 3 which is equal to notice that a 3 has a magnitude of 1 .3 and is pointing in the negative direction in both the x and y -axis.

02:58 So we have minus a minus 1 .3 times the cosine of 25 degrees i minus 1 .3 times the sign of 25 j.

03:15 A4 is pointing in the positive direction in the x direction and has no component in the y -axis.

03:22 So it's only 5 .1 i.

03:28 I, so all of these is in kilometers.

03:33 So a 5 has a magnitude of 1 .7 kilometers and it has a direction of 5 degrees east of north.

03:50 So notice that now the x component is 1 .7 times the sign of 5.

03:56 And both the x and y components are in the positive direction.

04:00 So we have 1 .7 times a sign of 5i plus 1 .7 times the cosine of 5.

04:11 J.

04:13 Then a6 has a magnitude of 7 .2 kilometers and it's pointing in the negative direction.

04:22 So it's minus 7 .2 cosine of 55, i minus 7 .2 times the sign of 55, i minus 7 .2 times the sign of 50.

04:34 A5j.

04:36 And a7, finally, it's pointing in the positive direction both in the x and y axes and has a magnitude of 2 .8...

Question

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