Change the chapter
You fly 32.0 km in a straight line in still air in the direction $35.0^\circ$ south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction $45.0^\circ$ south of west and then in a direction $45.0^\circ$ west of north. These are the components of the displacement along a different set of axes—one rotated $45^\circ$.
Question by OpenStax is licensed under CC BY 4.0.
Final Answer

a) South: $18.4 \textrm{ km}$, West $26.2 \textrm{ km}$

b) SW: $31.5 \textrm{ km}$, NW: $5.6 \textrm{ km}$

Note: in part B Shaun mistakenly switched units to meters instead of kilometers. The numbers are all correct, but please think "km", when you see "m".

Solution Video

OpenStax College Physics Solution, Chapter 3, Problem 21 (Problems & Exercises) (4:19)

Sign up to view this solution video!

View sample solution

Calculator Screenshots

OpenStax College Physics, Chapter 3, Problem 21 (PE) calculator screenshot 1
OpenStax College Physics, Chapter 3, Problem 21 (PE) calculator screenshot 2
Video Transcript

This is College Physics Answers with Shaun Dychko. So you fly 32.0 kilometers in a direction 35.0 degrees to the south of west. So that's this blue vector here and the question is asking us, if you had to get there by first going south and then going directly west, how far would you have to go south and then west. So that's the same as asking us what the components of this vector are and the y and the x directions. So the southerly amount of your trip, your first leg of your trip going directly south would be the distance multiplied by sine <i>theta</i>, that's 32.0 kilometers times sine 35 giving us 18.4 kilometers to the south. Then to the west, you would have to go <i>d</i> times cos <i>theta</i>. You'd have to go 32 kilometers times cos 35 because this leg here is the adjacent leg of this right triangle. So that's why we use cosine to find them. That's 26.2 kilometers. Then for part B, it says suppose you first went at an angle of 45 degrees south of west and then went at an angle of 45 degrees west of north, how far would you have to travel in each leg to reach the same destination. So that's the same as taking a rotated coordinate system and then finding the components along this new coordinate system. So we've rotated this coordinate system 45 degrees and we're first going to find out what is the <i>x</i> component on this rotated coordinate system which I've labeled <i>d subscript sw</i> or distance, or displacement, you know, south west component. Then here we have 45 degrees to the west of north. Here is north and then this is 45 degrees in there. This is the west of north component here. So first you'd have to go south west a distance of the total distance multiplied by cos <i>theta prime</i> and <i>theta prime</i> is the little bit that's in this new right triangle we've created with the axes of our rotated coordinated system and the vector in blue. So the <i>dsw</i> is the adjacent leg of that triangle and so we use cosine to find it and this angle would be 10 degrees because the axis here for <i>x prime</i> is at 45 degrees with respect to the original x axis and this vector is at 35 degrees with respect to that, the same original x axis, which leaves 10 degrees left over for this little bit in here, 45 minus 35 is 10. So we have 32 times cosine of 10 making 31.5 meters south west first of all. Then you'd have to go 45 degrees west of north and that would be 10 degrees with respect to the vector. Well, never mind that. We have 10 degrees between the rotated coordinate system in the vector. So we have, we're using the trigonometric function sine in order to find the opposite leg of this triangle in purple. We have 32 times sine of 10 makes 5.6 meters. It's always a good idea to check your work and here you could do that by saying let's take the square root of the sum of the squares of each of these components and we expect to end up with the original length of the displacement. Sure enough, square root of 31.5 squared plus 5.6 squared makes 32.0 meters which is this, length of this vector in blue.


Submitted by bryanlovell on Tue, 09/11/2018 - 06:09

Thank you for these videos. I do not have a strong math background and the textbooks don't really show all the work. Was able to get 100% on my last homework assignment. Well worth the money.

Submitted by kmonegro151 on Wed, 09/18/2019 - 20:02

why do you make the units for part B meters and not km

Submitted by ShaunDychko on Fri, 09/20/2019 - 10:03

Hello kmonegro151, thank you for noticing that mistake! I've made a note of it for other students. The numbers are correct, but the units should be "km".

In reply to by kmonegro151