Acceleration overcoming Velocity. Over what time frame?

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SUMMARY

The discussion focuses on a particle's motion under constant acceleration, specifically addressing the time it returns to its initial position (Xo) and its speed at that moment. The particle starts with an initial speed (Vo) in the positive x-direction while experiencing negative acceleration (a). The key equations involve kinematic principles, where the time of return (t2) can be derived from the relationship between initial speed, acceleration, and displacement.

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  • Study the kinematic equation for uniformly accelerated motion: X = Xo + Vo*t + (1/2)*a*t^2
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Homework Statement


A particle leaves its initial position Xo at time t=0, moving in the positive x direction with speed Vo but undergoing acceleration a in the negative x-direction. Find expression for (a) the time when it returns to Xo and (b) its speed when it passes that point



I'm having difficulting finding my second time t2. If I have this I can find the speed with V=(X-Xo)/(t-0). I'm trying to get into this course but am drawing a blank at this time. If someone could give me a few words in an explanation, I'd be greatly appreciative. It may just open my eyes.

Mike
 
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You can't get speed with V=(X-Xo)/(t-0). The speed is not constant, there is acceleration. How does speed depend on time under constant acceleration?
 
What have you tried to find t2?
 

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