Particle moving along a parabola

  • Thread starter Thread starter Peppino
  • Start date Start date
  • Tags Tags
    Parabola Particle
Click For Summary

Homework Help Overview

The discussion revolves around a particle moving along a parabola, with a displacement equation given as s = 0.5t². Participants are exploring the components of velocity and acceleration at specific points on the graph, as well as the implications of changing the x-component of velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to determine the direction of velocity and acceleration vectors, as well as the y-components of velocity and acceleration. Questions arise regarding the definitions of variables and the implications of the equations provided. Some participants express uncertainty about the conditions under which the particle's speed is considered constant.

Discussion Status

There is a mix of confirmations and corrections regarding the original poster's answers. Some participants have provided guidance on clarifying the definitions and relationships between variables, while others are questioning the completeness of the problem statement and the assumptions made about the motion of the particle.

Contextual Notes

There are indications of missing information, particularly regarding the relationship between the coordinates of the particle when moving along a parabola. Participants are also discussing the implications of the scalar equation provided and its relevance to the motion described.

Peppino
Messages
39
Reaction score
0
Can someone please check that everything I have done so far is correct?

A particle movies along a parabola with the displacement equation s = 0.5t2.
(a graph is shown)

::::Part I::::

Suppose x-component is s = Ct
i) indicate direction of velocity vector and acceleration at point R (arbitrary point on graph)
ii) Determine y-component of particles velocity
iii) Determine y-component of particles acceleration

Answers:

i) I suppose the velocity vector would be tangential to the graph and the acceleration vector would be parallel with the y-axis?
ii) since V = √(Vx2 + Vy2), and V = t, I found that the y - component is √(t2-C2t2).
iii) Since the acceleration is constant, is the y-component simply C?


:::Part II::::


Suppose instead that the x-component of the velocity is given by \frac{C}{\sqrt{1 + t^{2}}}

i) Show that the particle's speed is constant

I am unsure how to attempt Part II.

If you are able to answer or confirm my answers to any part of this question, thank you very much!
 
Last edited:
Physics news on Phys.org
Some help would be greatly appreciated!
 
We need some explanation. What is s? I do not see the graph.

ehild
 
For part (i) - you are right,
For part (ii) you have used s_x instead of V_x to determine V_y
For part (iii) - yes, the acceleration is constant, but it is not equal to C; but maybe try determining the total acceleration, and the acceleration in the x direction instead of differentiating...
- if you typo'd s=Ct; and it's actually V_x=Ct, then your answer to part (i) is wrong, part (ii) is right and you still need to determine part (iii) - it's a constant still, but not C

For part (II)
there isn't enoguh information - is the total motion of the particle still given by s=.5t^2?
Because in that case, the particles speed can't be constant - it's speed is, as you said, is equal to t, which is a variable.
Was there any other information?
 
s=0.5t^2 is a scalar equation. If s is displacement, the particle moves along a straight line, not along a parabola. If a particle moves along a parabola, you need to give the relation between its coordinates. Can you show the original text of the problem?

ehild
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
Replies
4
Views
2K
Replies
1
Views
1K
  • · Replies 40 ·
2
Replies
40
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K
Replies
11
Views
2K
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K