How Does Acceleration Affect Motion in the XY Plane?

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Homework Help Overview

The discussion revolves around the motion of a particle in the XY plane, specifically focusing on its velocity and displacement given initial conditions and acceleration. The problem involves determining the components of velocity and position over time, as well as exploring the distinction between displacement and distance traveled.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating velocity components and position vectors based on given acceleration. Questions arise about how to determine the distance traveled by the particle, with some suggesting the need to find an expression for speed as a function of time. Others explore integration of velocity components to find distance.

Discussion Status

The discussion is active, with participants providing hints and exploring different interpretations of how to calculate distance versus displacement. Some guidance has been offered regarding the integration of velocity to find distance, but no consensus has been reached on the final approach.

Contextual Notes

Participants are navigating the complexities of two-dimensional motion and the differences between distance and displacement, with some assumptions about the nature of the particle's movement and the mathematical relationships involved.

mohdhm
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1)
A particle starts form the origin at t=0 with an initial velocity having an x compoent and a y component. the particle moves in the xy plane with an x component of acceleration only


a)
Determine the components of velocity vector at any time and the total velocity vector at any time
Given Info: Vxi=20m/s, Vyi=-15m/s, Ax=4.0m/s^2, Ay=0.
Solved.
Answer: Vf=Vxi+Vyi = [(20+4.0t)(i hat symbol?) - 15(j hat symbol?)]m/s

b) calculate the velocity and speed of the particle at t=5?

Ok the tricky part... both velocity and speed have the same magnitude because with 2 dimensions you have to use the pythagorean theorem to add the magnitude of the x and y components. I also concluded that vector = speed but with theta(direction).

solved the problem for part B.

c) determine the x and y coordinate of the particle at any time t and the position vector of this time.

xf=vit=1/2at^2= (20t+2.0t^2)m
yf=vit=(-15t)m

so the position vector at any time t is [(20t+2.0t^2)i -15tj]m

Then the example question says.. let's test it out. and gives me some values

at t=5, x=150, y=-75

what is the magnitude of the displacement?

rf=sqrt[(150^2+(-75)^2]=170m

"note that this isnot the distance that the particle travels in this time! Can you determine the distance from the available data?

Now how do you determine the distance at t=5 ? do you just add the vector components of distance? If someone can please answer that last question, i could rest my mind :eek:
 
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mohdhm said:
Now how do you determine the distance at t=5 ? do you just add the vector components of distance?
I assume you mean add the components of the displacement. No, that won't do. Hint: Find an expression for the speed of the particle as a function of time.
 
so you find the intergral of Vx and Vy and add them up?

d(t)=Velocity * integral of 0 to 5 where f(x) is dt ?
 
Find an expression for the speed of the particle and integrate that from t = 0 to 5 seconds. What's the speed as a function of time?
 
finding the integral of vf=vi+at gives me d= vit+1/2at^2 and we just solve from there... using the absolute value...
i integrated the vf=vi+at and got the expected d=vit+1/2at^2 which gives me DISPLACEMENT

What is the distance, what is the solution of this problem.
 
Last edited:

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