Particle Distance/Velocity - Natural Logarithms

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Chase.
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Homework Statement



vDnRc.png


I'm going to post an image due to the complex syntax in the problem...

The Attempt at a Solution



I'm not going to lie... I really have no idea where to even begin with this problem. Because it says total distance traveled by the particle, I'm assuming that the total distance would be the distance equation multiplied by two; I'm not quite sure where velocity fits in however.

If I'm way off base, I'm still not looking for anyone to explicitly write out the answer for me. I just want to get headed in the right direction.
 
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First thing you have to determine is when the particle stops. Look at that equation: v(t) = v0 e^(-at). Does v(t) ever actually equal zero?

You need an infinite amount of time before the particle actually stops. In practical terms, the particle will be found to have slowed practically to a stop in a measurable amount of time. However, for this theoretical exercise, t has to be infinitely large before v becomes zero.

At this point, e^(-at) vanishes (becomes zero). Can you now find the value of s at this point?
 
Chase, the ball is back in your court. Show us what you have done, using the hints given by Courious3141 (this is, in fact, a very easy problem and Courious3141 gave very goo hint) or this thread will be deleted.
 
HallsofIvy said:
Chase, the ball is back in your court. Show us what you have done, using the hints given by Courious3141 (this is, in fact, a very easy problem and Courious3141 gave very goo hint) or this thread will be deleted.
I'm not sure why this response was so aggressive.

Anyway, thanks Curious! I figured it out. I had initially considered the v0/a solution but discounted it because I didn't think I could input v0 in the submission box.
 
Chase. said:
I'm not sure why this response was so aggressive.

Anyway, thanks Curious! I figured it out. I had initially considered the v0/a solution but discounted it because I didn't think I could input v0 in the submission box.

Glad you got it.:smile: