Finding Acceleration in Two-Dimensional Kinematics

  • Thread starter Thread starter indietro
  • Start date Start date
  • Tags Tags
    Kinematics
Click For Summary

Homework Help Overview

The discussion revolves around a problem in two-dimensional kinematics, specifically focusing on determining the acceleration of a flying saucer given its positions and velocities. The original poster expresses confusion regarding the absence of time in the problem and seeks alternative methods to find acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the chain rule to relate velocity and position without time. Questions arise about the interpretation of vectors in the context of the trajectory and the meaning of the position vector.

Discussion Status

Participants are actively exploring various interpretations of the problem, with some suggesting methods to relate different graphs and vectors. There is no explicit consensus, but guidance is being offered regarding the use of equations versus graphical methods.

Contextual Notes

The problem is constrained by the lack of time information, and participants are attempting to navigate the implications of this absence while discussing the provided graphical representations of velocity.

indietro
Messages
46
Reaction score
0

Homework Statement


a flying saucer manueverign with a constant acceleration is observed with the positions and velocities shown below. what is the saucer's acceleration?

Homework Equations


a = [tex]\Delta[/tex]v / t

The Attempt at a Solution


so a time is not given.. is there another way to find acceleration?
** actually if someone knows a site that gives a clear and concise tutorial on two-dimensional kinematics that would be great :). I know there is one attached to this site.. but i do better with actual examples and more visual.
 

Attachments

  • 7.jpg
    7.jpg
    3.1 KB · Views: 426
Physics news on Phys.org
indietro said:
a flying saucer manueverign with a constant acceleration is observed with the positions and velocities shown below. what is the saucer's acceleration?

so a time is not given.. is there another way to find acceleration?

Hi indietro! :smile:

(I can't see the picture yet, but …)

The standard trick for finding dv/dt from v and s without involving t is to use the chain rule …

dv/dt = dv/dx dx/dt = v dv/dx :wink:
 
oo ok for the picture it is a x-y graph that shows a point at (0.0) with a vertical vector (200[tex]\hat{j}[/tex] m/s) and a second point at (2000, 1000) with a south-east vector (200[tex]\hat{i}[/tex] - 100[tex]\hat{j}[/tex] m/s)

for the chain rule: is the v the final velocity?
 
v = dx/dt
 
so i have a question: for an x-y graph showing the trajectory, what does [tex]\vec{r}[/tex] (starts at origin and goes to a point on the trajectory) tell me? the velocity at that point? or does it only tell me the direction of velocity at that point?
 
sorry but I am really confused as how to relate an xy-graph, a vx graph, a vy graph and acceleration. Like what does each tell me, how can i find position after a certain time, how can i find acceleration?
 
Hi indietro! :smile:

(just got up :zzz: …)
indietro said:
so i have a question: for an x-y graph showing the trajectory, what does [tex]\vec{r}[/tex] (starts at origin and goes to a point on the trajectory) tell me? the velocity at that point? or does it only tell me the direction of velocity at that point?

r only tells you the position, and the direction of the tangent of the trajectory tells you the direction of the velocity.
indietro said:
sorry but I am really confused as how to relate an xy-graph, a vx graph, a vy graph and acceleration. Like what does each tell me, how can i find position after a certain time, how can i find acceleration?

Why are you using a graph? Does the question tell you to?

If it doesn't, then forget graphs, and just use equations. :smile:
 
yes the question gives me all the information in the form of an vx graph and vy graph ...:(
 
indietro said:
yes the question gives me all the information in the form of an vx graph and vy graph ...:(

hmm … I still can't see your picture.

I'll have to wait until I can see it.

(I've reported it, so hopefully it'll come up soon :redface:)
 
  • #10
oooh, I see it now! :-p

ok, that isn't a graph (a graph would be a continuous curve) …

it's just a diagram, defining the two velocities in a picture instead of in words.

So you don't have to use a "graph method" …

just write the velocities as 200j and 200i - 100j, and carry on from there. :smile:
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
4K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
15K
  • · Replies 33 ·
2
Replies
33
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K