# Simple, yet incredibly frustrating, net displacement vector problem [closed]

This is for MasteringPhysics and I've wasted 7 out of 9 tries and an hour trying every possible solution to this problem. Sig figs are not an issue and I'll finish my gripe after I show you the problem.

While vacationing in the mountains you do some hiking. In the morning, your displacement is $S_{morning}= (2500 m, east) + (4000 m, north) + (200 m , vertical)$. After lunch, your displacement is $S_{afternoon}= (1600 m , west) + (2000 m , north) - (300 m , vertical)$. What is the magnitude of your net displacement for the day?

What gets me about this problem is that I can't seem to get an answer that MasteringPhysics will accept. Displacement, to my knowledge is change in position. I start at the origin, and I end up 1600 meters west of it, 2000 meters north of it, and -300 meters down from it. I've changed reference points and tried nearly every combination of the given numbers. Maybe I'm missing something, or maybe someone coded this question incorrectly, but I'll show my thought process below and let you be the judge of that.

Assuming that my starting position is the origin, my net displacement for the day would simply be the given displacement in the afternoon, assuming I didn't return to my starting location. Calculating the magnitude if $S_{afternoon}$ using the distance formula gives me 2600 meters, assuming two sig figs. Just in case these did matter, I also tried 3000 meters. No go.

Then I approached the problem with the idea that $S_{morning}$ would be the starting reference point and came up with the vector (4100, -2000, -500), the magnitude of which was 4600 meters. It didn't work. At this point I just started trying random combinations, none of which worked.

My answers thus far have been: $$4600, 2600, 6100, 2579, 4700, 6500, 3000.$$

My units are meters (m) and I'm at my wit's end.

This is somewhat just a follow up to the answer from user1583209, which has the answer for your question. But I think I see the cause of all your troubles, and I want to address that.

The displacement of the day means the total displacement of everything that happened that day. This "total" is the sum!

You have two displacements during this day and you must add them: $$s_{day} =s_{morning}+s_{afternoon}$$

Not subtract them as in your latter try or only consider one of them as in your former try. These are two independent displacements that happen from whatever their current position was.

It is like pushing on a chair twice: the total displacement is both pushes added up.

And note - you might have noticed this, but I'm not sure - that they are trying to trick you with East and West. A displacement towards West is of course just a negative displacement towards East. So these can be added together with fitting signs.

• Could mention that when adding the displacements one adds all coordinate directions independently, i.e. displacements in North-South directions go together as do East-West displacements as do up-down in vertical directions. And watch out that when you have e.g. a displacement in east direction and one in west direction you are actually retracing, so need to subtract the numbers. – user1583209 Feb 6 '17 at 0:51
• That you can add those displacements independently is essentially due to the fact that you can do the displacements in any order ("operators commute"), i.e. if you do the afternoon hike before the morning hike you would still end up at the same spot. – user1583209 Feb 6 '17 at 0:57
• @user1583209, good points, I've added some to the answer. – Steeven Feb 6 '17 at 1:00