Constant acceleration motion problem

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Homework Help Overview

The problem involves a particle moving along a straight line with constant acceleration, changing direction at t = 4 seconds, and reaching a specific position and velocity at t = 10 seconds. The goal is to determine the position of the particle at the moment it changes direction.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster expresses confusion regarding the definition of variables and the application of equations of motion. They consider the initial conditions and the relationships between position, velocity, and acceleration.
  • One participant questions the derivation of a specific equation used in the solution process.
  • Another participant clarifies their method of solving the equations by subtraction, which leads to a specific value for acceleration.

Discussion Status

Some participants have provided guidance and clarification on the problem-solving process. The original poster has indicated they found a solution independently, while others are still engaging with similar questions related to the topic.

Contextual Notes

The original poster mentions a quiz deadline, indicating a time constraint that may influence the urgency of the discussion. There are references to a shared educational experience among participants, suggesting a collaborative environment.

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Homework Statement


A particle moving along a straight line with constant acceleration starts its motion at t = 0. The particle is observed to change direction at t=4s, and when t=10s it reaches the position of -2 m with a velocity of -2.4 m/s measured from a chosen reference frame. Find the position of the particle when it changes direction of motion.

Answer: 5.2 m


Homework Equations



x = x0 + v0*t + (1/2)at^2
v = v0 + at

and possibly: v^2 = v0^2 + 2a(x-x0)

The Attempt at a Solution



I'm not really sure how to go about doing this problem. I'm getting a little confused on the defining variables part even -- this is what I THINK they should be defined as but I'm not sure:
x = -2 m at t=10
x0 = ?
v = 0 m/s at t=4s, -2.4 m/s at t=10
v0 = ?
a = ?

It seems like I'm missing 3 things. I'm thinking either v0 or x0 is 0, but I'm not sure which of those is 0.

I think I'll only be using the first 2 equations since time is involved, and that's really as far as I got on this problem. :( Any help is appreciated, I have a quiz that includes this problem due tomorrow at 1:25 PM eastern time.
 
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Someone can close this -- I figured it out on my own. Used systems to solve this one per a friend's tip. Solution is below for anyone interested:

0 = v0 + 4a
-2.4 = v0 + 10a
==============
2.4 = -6a
a = 2.4/-6 = -0.4

0 = v0 + 4 * -0.4
v0 = 1.6

-2 = x0 + 1.6 * 10 + .5 * -0.4 * 100
x0 = 2

x = 2 + 1.6 * 4 + .5 * -0.4 * 16
x = 5.2 m
 
How did you get the -6a?
 
anon_tss said:
How did you get the -6a?

I just subtracted the two equations. That's why the 2.4 is positive on the lefthand side and the v0 disappears.

Also, this thread is OLD and has been solved already. Someone can close it as I said in the previous post
 
I got a similar question on a quiz due tomorrow, Chang ftw
 
anon_tss said:
I got a similar question on a quiz due tomorrow, Chang ftw

Haha nice, you're a VT student too? I had Chang a couple semesters ago...I thought he stopped teaching?
 
Ya, second semester at tech. Chang still teaches, I was pretty lucky to get him haha
 
anon_tss said:
Ya, second semester at tech. Chang still teaches, I was pretty lucky to get him haha

Yeah, honestly, all the Physics professors EXCEPT for Chang are horrendously bad at VT. Enjoy 2306 next semester...it's a fun time.
 
Ya I heard, thanks for the help. I have to figure out the rest of the stuff haha
 
  • #10
anon_tss said:
Ya I heard, thanks for the help. I have to figure out the rest of the stuff haha

No problem
 

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