Displacement from an Acceleration Time Graph

Click For Summary
The discussion focuses on calculating displacement from an acceleration time graph using kinematic equations. The user has determined the time at which velocity returns to 54 km/h but struggles with finding the correct position. They attempted various equations, including xo = vot + 1/2at² and x = xo + vt, but arrived at an incorrect displacement of 270m instead of the expected 178.8m. Suggestions include setting up acceleration functions for different regions and integrating them to find displacement. The conversation emphasizes the importance of accurately representing the graph and showing detailed attempts for better understanding.
CGI
Messages
74
Reaction score
1

Homework Statement


upload_2016-2-5_1-42-37.png

Homework Equations


Kinematic equations
t = 18s when the velocity is back to 54km/h or 15m/s
at t = 2, V = 13.3 m/s
at t = 4.5, velocity is 1.5 m/s
at t = 1, V = 15m/s

The Attempt at a Solution


I have solved the first part and I have the time at which it is 54km/h again, but now I'm trying to figure out
the position. I think I could make a v-t graph and find the area under that curve, but it seems like that would be too hard to accurately represent the graph. Is there any other way to find displacement? I've tried
xo = vot + 1/2at2 but that doesn't seem to help
I also tried
x = xo + vt, but I get 270m as an answer. The correct answer is 178.8, but I just want to understand how to do it. Any help would be much appreciated!
 

Attachments

  • 20160204_200131.jpg
    20160204_200131.jpg
    19.8 KB · Views: 662
Physics news on Phys.org
What you can try is set up a(t) functions for the two regions and integrate the functions twice wrt time.
 
You'll have to show your attempt(s) in detail.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
25
Views
2K
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
7
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
1
Views
2K