^{5}- 25t

^{3}+60t.

^{ }

A particle moves on the x-axis in such a way that its position at time t is given by x(t)=3t^{5} - 25t^{3} +60t.^{
}

For what values of t is the particle moving to the left?

A) -2<t<1 only

B) -2<t<1 and 1<t<2

C) -1<t<1 and t>2

D) 1<t<2 only

E) t<-2, -1<t<1, and t>2

Tutors, sign in to answer this question.

1. A particle is moving to the left when its velocity is negative.

2. The velocity is the derivative of the displacement versus time function.

v(t) = s'(t) =15t^{4}-75t^{2}+60 <= 0

This function can be factored by first factoring out the 15 to give 15(t^{4}-5t^{2}+4) which in turn can be factored

15(t^{2} - 4)(t^{2} - 1). The two terms can be further factored because each is the difference of two squares.

15(t-2)(t+2)(t-1)(t+1) < 0. Kenneth has provided the rest of the solution. I just wanted to help you factor the polynomial.

The correct answer is E because x'(t) = 15(t-2)(t+2)(t-1)(t+1)

and this velocity polynomial is positive on three intervals: (-infinity,-2) (-1,1) (2,infinity)

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