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Bensky
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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.