Position vector of a moving particle (calc based intro)

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Homework Help Overview

The problem involves determining the position vector of a particle moving in a horizontal xy-plane, given its acceleration, initial position, and velocity at a specific time. The context is rooted in kinematics and calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration, velocity, and position, particularly in the context of non-constant acceleration. There are questions about the correct application of kinematic equations and the role of calculus in finding position from acceleration.

Discussion Status

Some participants have offered insights into the need for integration and differentiation when dealing with varying acceleration. There is an exploration of how to approach the problem using both standard kinematic equations and calculus, with no explicit consensus reached on the best method.

Contextual Notes

Participants are considering the implications of non-constant acceleration and the potential need for integrating or differentiating in their problem-solving approach. There is also mention of the challenge of memorizing results versus understanding derivations.

MurdocJensen
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Homework Statement



Reviewing for a midterm. I have this problem I can't seem to nail. Want to try it?

"The acceleration of a particle moving only in a horizontal xy-plane is given by a=<(3t)i,(4t)j>, where 'a' is in m/s^2 and 't' is in seconds. At t=0, the position vector r=<(20)i,(40)j> locates the particle which then has the velocity vector v=<(5)i,(2)j>. At t=4s, what is the particle's position vector in unit-vector notation?"

to clarify units: v is in m/s, a is in m/s^2, position is in meters.


Answer: r=<72i,90.2j>


Homework Equations



The standard kinematics equations, I'm assuming. If calc is involved, I wouldn't know where to implement it.

The Attempt at a Solution



I think it's the acceleration at t=4 seconds that's tripping me up. When I plug in for position in the 'x' or 'y' direction I am using t=4 to multiply with the original acceleration components. Am I not finding acceleration correctly? And wouldn't it stand that if I multiplied an acceleration component by a time I would get a velocity?
 
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if you were given x(t) , how would you find a(t) ?
yes, using calculus.
this is the "inverse" question ; they give a(t) ... not constant ... and want x(t)
... so use the "inverse" operation.
 
lightgrav said:
if you were given x(t) , how would you find a(t) ?
yes, using calculus.
this is the "inverse" question ; they give a(t) ... not constant ... and want x(t)
... so use the "inverse" operation.

Yes, if you want to find acceleration from position you would take two derivatives right?

So if you want to find position from acceleration you would try to "undo" derivatives (don't forget the constants).
 
This is very helpful, thank you. Now, in general, would I consider integrating and/or differentiating in a situation where my acceleration isn't constant?

I ask because I want to know if there is a way to do this problem with your 'plain-vanilla' kinematics equations.

Actually, let me restate. When should I consider integration and or differentiation in a physics problem?
 
Here, the acceleration increases at a constant rate (da/dt = jerk = const),
so you _can_ find the average acceleration in a simple route , here .
... if you want to, you can sprinkle a " + 1/6 j t^3" term onto the vanilla
... and dip it in fudge by adding " + 1/24 d t^4 " , etc ... where to stop ?!?

well, you got those "plain-vanilla" kinematic equations via time-derivatives of location .
The better you understand things, the less you need to memorize;
but some things are SO often encountered (like constant accel)
and some derivations are TOO nasty of algebra to do on an Exam.
. . . so, you should memorize a few results (2 per chapter?)
 

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