A particle moves along the x-axis so that at any time t>0 (or equal to 0), its position is given by x(t)=t^3-3t^2-9t+1. For what values of t is the particle at rest?
A) No values
B) 1 only
C) 3 only
D) 5 only
E) 1 and 3
Any explanation would be greatly appreciated! Thank you!
A) No values
B) 1 only
C) 3 only
D) 5 only
E) 1 and 3
Any explanation would be greatly appreciated! Thank you!
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I wish I could help. I did study all this one time. I forget. If you could draw the graph this equation describes it would help - presumably a squiggly line that crosses the x axis at certain points. I think you need a result something like (t+p)(t+q)(t+r). Then you solve the equation for when x(t)=0 and get t= -p, t= -q, t= -r. These are the values at which the graph crosses the x axis (ie when x(t) = 0), because (p-p)=0 and will make x(t) = 0, and similarly for (q-q) and (r-r). Try it: put t= -p in the equation: (t+p)(t+q)(t+r). You get zero. Same for the others. So these are the 3 values of t that make the x(t) = 0. I am assuming that when x(t) = 0, the particle is at rest. But I could be wrong. However, if you use the calculator below, you will find 3 different values. If the calculator is right, and I put in the right values, then the answer is C) I think.