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Displacement vs Time problem 
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#1
Sep111, 04:22 PM

P: 1

1. The problem statement, all variables and given/known data
The position of a particle moving along the xaxis depends on time as x(t) = 100 t^2  8 t^3 where x is in m and t in s. At what time does the particle reach its maximum positive x position? What is the length of the path covered by the particle between t = 0.0000 s and t = 10.833 s? What is the displacement of the particle between t = 0.0000 s and t = 10.8333 s? 2. Relevant equations x(t) = 100 t^2  8 t^3 3. The attempt at a solution I haven't had any attempts that have led me anywhere, really. This is the first time I've encountered this type of problem. 


#2
Sep111, 06:11 PM

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P: 6,202

The length of the path would be the arc length between the two points.
The displacement would be the straight line distance between the two points. 


#3
Sep111, 06:19 PM

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PF Gold
P: 1,954

Hello bmadkins,
Welcome to Physics Forums! But perhaps a more intuitive guide (which is the same thing applied to this problem), is to consider that at some moment in time, the particle is moving away from its original position. Eventually, it slows down, momentarily stops, and then goes back in the other direction toward the original position. So at what point was it farthest away from the original position? At the instant it momentarily stopped  at the instant it "turned around." At that point its instantaneous velocity is zero. So part of this problem is figuring out at what time (or times) the particle's velocity is zero. So to start, you might want to ask yourself, "what is a particle's velocity (as a function of time) if you already know what its position is (also as a function of time)?" (And I don't mean s = vt, since the particle is not moving at a constant velocity. And I don't mean v = at or s = ½at^{2} either, since the acceleration is not necessarily uniform. There is a more fundamental relationship between position and velocity.) 


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