The Time of Return for a Particle Undergoing Negative Acceleration

In summary, the conversation is discussing how to find the time when a particle returns to its initial position after starting at x_0 with a velocity of v_0 and acceleration in the negative x direction. The suggested equations are v=v_0-at and x=x_0+v_0*t, but it is then pointed out that these are for uniform motion and a quadratic equation using x = Vo*t + .5*a*t^2 is needed. This will result in two solutions, one being t=0 and the other being greater than zero.
  • #1
estie
14
0

Homework Statement



A particle leaves its initial position x_0 at time t=0 moving in the positive x direction with speed v_0 but undergoing acceleration of magnitude a in the negative x direction. Find expressions for (a) the time when it returns to the position x_0.

Homework Equations



v=v_0-at
x=x_0+v_0*t

The Attempt at a Solution



i know that v=v_0, but that gives me a time of 0, which doesn't make sense. I'm also not supposed to use x_0 in the solution. otherwise, i think the solution should be (v-v_0)/-a.
 
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  • #2
x=x_0+v_0*t
This formula is for uniform motion (zero acceleration).
You need to use x = Vo*t + .5*a*t^2.
When you set x = 0, you will have a quadratic equation with two solutions. One will be t=0 of course and the other greater than zero.
 
  • #3

Homework Equations



v=v_0-at
x=x_0+v_0*t
By the way estie, if you wish to use TeX-style notation on this forum you can wrap [ tex] [/tex] tags around your equations so it'll look more like this:
[tex]x=x_0+v_{0}t
[/tex]

Welcome to PF!
 

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