Particle Motion with Constant Acceleration

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SUMMARY

The discussion focuses on solving a physics problem involving a particle moving in the xy plane with constant acceleration. The initial position is given as rvec1 = (3.7 m)i + (3.4 m)j, and after 3 seconds, the position changes to rvec2 = (9 m)i − (1.9 m)j, with the final velocity vvec2 = (5.4 m/s)i − (6.5 m/s)j. Participants clarify the use of kinematic equations, specifically s = vt + (1/2)at², to derive the initial velocity vvec1, acceleration, and the functions for velocity and position over time. The discussion emphasizes the importance of understanding vector addition and integration in solving for velocity and position functions.

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  • Basic understanding of motion in two dimensions
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Loppyfoot
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Homework Statement


A particle moves in the xy plane with constant acceleration. At t = 0 the particle is at rvec1 = (3.7 m)i + (3.4 m)j, with velocity vvec1. At t = 3 s, the particle has moved to rvec2 = (9 m)i − (1.9 m)j and its velocity has changed to vvec2 = (5.4 m/s)i − (6.5 m/s)j. (a) Find vvec1.(b) What is the acceleration of the particle?(c) What is the velocity of the particle as a function of time?(d) What is the position vector of the particle as a function of time?

I need some guidance. I tried using the average velocity formula for (a), but it doesn't seem to be working for me. I tried doing 9-3.7 / 3. for the i vector. And likewise, for the j vector (-1.9-3.4)/3. I get 1.76i-1.76j, and it isn't correct.

I need some guidance on the others too. Thanks guys.
 
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Jeez, I cannot seem to get this! I have been trying to get it all night.
 
Hi Loppyfoot! :smile:

(use bold for vectors :wink:)
Loppyfoot said:
… I tried using the average velocity formula for (a), but it doesn't seem to be working for me.

Distance velocity and acceleration are all vectors, and so they add like vectors.

Try s = vt + (1/2)at2 :wink:
 
How should I implement that equation into part a), if I do not know the acceleration vectors yet?
 
Part of your problem description doesn't make sense to me.
Edit: Now it does make sense.
Loppyfoot said:
A particle moves in the xy plane with constant acceleration. At t = 0 the particle is at rvec1 = (3.7 m)i + (3.4 m)j, with velocity vvec1. At t = 3 s, the particle has moved to rvec2 = (9 m)i − (1.9 m)j and its velocity has changed to vvec2 = (5.4 m/s)i − (6.5 m/s)j.
Edit: Ignore the following.
This description says that the particle moves with constant acceleration. Later it says that the velocity has changed to ... If the acceleration is constant, the velocity can't change, since acceleration is the instantaneous rate of change of velocity with respect to time.
 
Last edited:
If the acceleration is constant, doesn't that mean that the velocity is changing at a constant rate?
 
Never mind. I take back what I said. I was thinking zero acceleration, not constant acceleration.
 
I think I figured it out. I applied one of the 4 kinematic equations, and then I should be able to get the rest from there.

EDIT: I got a and b, but how would I go about getting c and d?
 
For c, if you got b, use it to get the velocity. a = dv/dt, so you can integrate what you have for a to get v as a function of t. You'll get a constant (vector) of integration, but you know v(3), so should be able to figure out the constant.

For d, do essentially the same thing: v = ds/dt. Integrate that to get s and use the given information about s(0) to figure out this constant (vector) of integration. Does that make sense?
 

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