For a particle motion question, I am getting a negative time. What do I make of this?

OllieOpps

New member
Joined
Dec 3, 2019
Messages
2
Hello! So the function is s(t)=t3-3t2-9t-1. The 1st derivative or v(t) is 3t2-6t-9. When I solve for zeros, I get t= 3 and -1. What does this negative time do to my problem? Do I disregard it and use only 3?(I wouldn't make sense to me, since the negative time indicates a moment in which a change may occur?) Thank you!
 
You have found the zeroes of the function's first derivative correctly. Those zeroes indicate the times, relative to your base time of zero, when the direction of motion changes. (I am guessing that s stands for distance from a base point.) There is no mathematical reason why the negative time should be ecxluded: it merely represents one unit of time before the time you called time zero.

Now negative times may not apply to the specific problem that you are attacking through mathematics. How in the world can we determine that if you do not give us the exact and complete problem as we ask you to do in the submission guidelines? So t = -1 may be a valid answer or may not be; it depends not on mathematics but on the meaning of the variables in your specific problem, which we don't know.
 
Please tell us the full problem. Does it say that t represents time? Just because the letter "t" is used in an equation does not necessarily mean it represents time. Or a negative time might be perfectly reasonable, referring a time before to some arbitrary t= 0.
 
Top