Change the chapter
Question

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: 2.50 km $45.0^\circ$ north of west; then 4.70 km $60.0^\circ$ south of east; then 1.30 km $25.0^\circ$ south of west; then 5.10 km straight east; then 1.70 km $5.00^\circ$ east of north; then 7.20 km $55.0^\circ$ south of west; and finally 2.80 km $10.0^\circ$ north of east. What is his final position relative to the island?

Question by OpenStax is licensed under CC BY 4.0.
Final Answer

$7.34 \textrm{ km, } 63.5^\circ \textrm{ S of E}$

Solution Video

OpenStax College Physics Solution, Chapter 3, Problem 23 (Problems & Exercises) (8:56)

Sign up to view this solution video!

View sample solution

Calculator Screenshots

OpenStax College Physics, Chapter 3, Problem 23 (PE) calculator screenshot 1
OpenStax College Physics, Chapter 3, Problem 23 (PE) calculator screenshot 2
OpenStax College Physics, Chapter 3, Problem 23 (PE) calculator screenshot 3
Video Transcript
This is College Physics Answers with Shaun Dychko. Gilligan's raft is being blown in a lot of different directions after he sets off from the island which we take to be at the origin of this coordinate system, that big black dot in the middle. The first leg of his journey is 2.5 kilometers, 45 degrees north of west and I've drawn that in blue and labeled it vector A. Then he's blown 60 degrees south of east, a distance of 4.7 kilometers and he ends up here. Then he's blown 25 degrees south of west, 1.3 kilometers along vector C; then he goes 5.1 kilometers straight east, vector D; then he goes 1.7 kilometers, 5 degrees east of north, that's along here; and then 7.2 kilometers, 55 degrees south of west; and then finally 2.8 kilometers vector F, 10 degrees north of east and he ends up here. So the final position of his raft is this vector drawn in red, this resultant which goes from the origin to the very last head of the last vector. So each vector has been placed head to tail in order to add them altogether. The resultant goes from the origin to the head of the last vector. Okay. So first we're going to find the x component of this resultant by adding together the x component of each of these vectors. The only, well there's two things that are tricky with this question. Number one, it's just a lot of work because there's so many vectors and so it's easy to just make careless mistakes of copying numbers and so on. The second thing that's tricky is that this particular angle here is given as 5 degrees east of north which is unusual. Most every other, every other angle has been given with respect to a horizontal axis. It's either north of east, north of west, or south of east, south of west. But this one's different and that it's been given with respect to a vertical axis, it's east of north. So that's going to change which trigonometric function we use for finding the components of vector E. To find the x component of vector E we'll be using sine. To find the y component of vector E, we will be using cosine which is different from every other vector. Okay. So if you've made a mistake and you're not sure where the mistake is, check that out first. So, here we go. Vector A is in the negative direction for its x component, that's 2.5 kilometers times cosine of 45, cosine being the adjacent leg of this triangle here. Then we have plus 4.7 kilometers, that's positive because vector B has an x component pointing to the right and so we're looking at vector B now, which is this triangle here, and 4.7 times cosine of 60. Then minus 1.3 kilometers times cos 25 for the x component of vector C. So 25 is the angle in there. This is the triangle we're considering and we're looking for the adjacent leg of that triangle and it's pointing to the left so it's negative and that's 1.3 times cos 25. Then we have no, oh sorry, the x component of D, yeah it's all in the x direction and it's positive. So we just write 5.1, you could write times cosine of zero if you like but that doesn't make any difference 'cause cosine of zero is the number one. Then finally, we get to the tricky E vector and that x component is positive. It's the opposite leg of this triangle and so to find the opposite, we multiply the angle by -- sorry, we take the sine of the angle and multiply that by the hypotenuse. So that's 1.7 times sine of five, then subtract from that the x component of vector E. Its x component is the cosine of 55 multiplied by 7.2. Then finally, the x component of vector F which is cosine of 10 multiplied by 2.8, and that gives you 3.2799 kilometers. That's positive and that's the x component of the resultant. Then a similar sort of thing for the y component now. We're using sines in place of cosines that we had before, or in the case of the E vector, where is that -- we're using cosine instead of sine that we had before an we're choosing positives and negatives depending on whether the y components are going up or down. Let's get rid of some of our highlighting here and let's do it all again for the y component. So y component of A 2.5 times sine 45 and then minus 4.7 times sine of 60, we're looking at vector B now, 60 being this angle in here and finding the opposite leg so that means use the sine of the angle, sine of 60 multiplied by the hypotenuse 4.7. Then minus 1.3 times sine 25, looking at this triangle now, finding its opposite leg. The y component is directed down that's why there is a minus there, right there. Then, plus the y component of this vector. This vector did not appear at all, there's no 5.1 in this work at all because it has no y component for vector D. Vector E has a y component of cosine of 5 times 1.7, where's -- that's right there, good, then the vector E, we need to find it. Just noticed that my letters are a bit off here, aren't they? There's actually another -- oh my goodness! Well, I don't really feel like re-starting the video if I can help it. So let's just call this G, and this is F, and the labels don't really matter. So E, F, and G. Great! Yeah, so we're looking at vector F now, this green one that's 55 degrees south of west and 7.2 kilometers long. That y component is in the downward direction so we have a minus there. Then we have finally 2.8 kilometers vector G multiplied by the sine of 10 and that's positive because that y component is upwards. That gives us negative 6.5701 kilometers and the length of this resultant is going to be the square root of each component squared added together. So that's the square root of 3.2799 kilometers squared plus negative 6.5701 kilometers squared, giving us 7.34 kilometers is the length of the resultant. Then to find the position, we take the inverse tangent of the size of the y component of the resultant divided by the size of the x component of the resultant. So we're looking at this being the x component of the resultant and then this being the y component of the resultant and the angle we're finding is this one in here. So our answer will be south of east as we take the inverse tangent of the opposite divided by the adjacent. So the inverse tangent of 6.501 kilometers divided by 3.2799 kilometers and our final answer then is that the position of Gilligan's raft is 7.34 kilometers, 63.5 degrees south of east.