Vector displacement problem trouble

AI Thread Summary
The discussion centers around a vector displacement problem involving hiking displacements in the mountains. The user struggles to find the correct method to determine the change in altitude from the starting point, initially misinterpreting the problem by focusing on overall displacement rather than just the vertical changes. It is clarified that only the vertical components should be considered for calculating the altitude difference. Additionally, the user is advised to project displacements onto a single plane to find net x and y components, while the z component remains known. The conversation emphasizes the importance of correctly interpreting vector components to solve the problem accurately.
M1ZeN
Messages
17
Reaction score
0

Homework Statement


While vacationing in the mountains, you do some hiking. In the morning, your displacement is S-morning = (1600 m, east) + (3300 m, north) + (160 m, vertical). After lunch your displacement is S-afternoon = (1800 m, west) + (2000 m, north) - (350 m, vertical). At the end of the day, how much higher or lower (in m) are compared to your starting point in the morning?


Homework Equations


A^2 + B^2 = C^2


The Attempt at a Solution


I can't find any other relevant equations that is in the chapter and no examples are given with a third displacement value.

My attempt in the problem using the same method for both the morning values and after lunch values:

(1600)^2 + (3300)^2 = C^2
c = 3667.4 m

then...

(3667.4)^2 + (160)^2 = C^2
c = 3670.9 m

I used the same method for the afternoon values, found the difference between the two (1003.1 m), and it was the incorrect answer.

Thanks,
Daniel
 
Physics news on Phys.org
In the morning, how much up you go?
In the afternoon, from the above point how much down you go?
So what is the final depth from the starting point?
The other distances are just distractions.
 
M1ZeN said:

Homework Statement


While vacationing in the mountains, you do some hiking. In the morning, your displacement is S-morning = (1600 m, east) + (3300 m, north) + (160 m, vertical). After lunch your displacement is S-afternoon = (1800 m, west) + (2000 m, north) - (350 m, vertical). At the end of the day, how much higher or lower (in m) are compared to your starting point in the morning?


Homework Equations


A^2 + B^2 = C^2


The Attempt at a Solution


I can't find any other relevant equations that is in the chapter and no examples are given with a third displacement value.

My attempt in the problem using the same method for both the morning values and after lunch values:

(1600)^2 + (3300)^2 = C^2
c = 3667.4 m

then...

(3667.4)^2 + (160)^2 = C^2
c = 3670.9 m

I used the same method for the afternoon values, found the difference between the two (1003.1 m), and it was the incorrect answer.

Thanks,
Daniel
Daniel I think you are misreading the problem, they ask for the change in altitude from your starting point. Assuming that is 0m you just need to add or subtract the vertical distances they give you using say 0m as your starting point.

Since they give you the changes in vector form you can call the last piece the z axis or K-hat values of the vector and ignore the other dimensions.

Hope that helps

If it doesn't then this should be your answer
0+ 160m - 350m = -190m so your ending altitude would be 190 m less then your starting point
 
I found the correct answer for that part. Now the second part has asked me to: What is the magnitude of your net displacement (in km) for the day?

I went back to the method I used and just added the separate values for both morning and afternoon. It was the incorrect answer but I figured what I did was reasonable enough to try to find an answer.
 
Project every displacement on the single plane, then find the net x, y component. z component is known. Find net displacement.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top