Question

A particle is moving along a horizontal line according to the equation s(t)=t^4-t^3-t-1. In what time interval is the particle moving to the negative direction?

Please help me factor this. I'm not good at factoring when the exponent reaches 4. :(

Answer 1

this will not factor - if you equate it to 0 there are two values of t -0.618 and 1.618
at time = 0 it is at a displacement of -1 from the origin

Answer 2

ow. I'm confused. I didn't really get it, but thanks for the reply. :)

Answer 3

ok

Answer 4

if you take the derivative, you will be able to "see" if move.

for example; the position function has a derivative which tells us the velocity at any given time.

if the velocity is + its moving forwards; and if - its moving backwards

Answer 5

\[\frac{d}{dt}(t^4-t^3-t-1)=4t^3- 3t^2-1 \]

4t^3- 3t^2-1 ; is what we need to find the zeros for i believe

Answer 6

That becomes v(t)=4t^3-3t^2-1...I'm suppose to make a table of signs out of this right?

Answer 7

yes

Answer 8

I factored it just now and it becomes v(t)=(t-1)(4t^2+t+1).. Now I'm stumped. I'm quite doubtful if I should use the quadratic formula on (4t^2+t+1).

Answer 9

you can use the quad formula on any quadratic; the results might be an imaginary number, but it doesnt hurt to try

Answer 10

its quite possible that the v(t) has one "hump" and one inflection; since inflections dont change direction its fine

Answer 11

lets test it out at t=1 tho :) since i believe that will be our only extrema

Answer 12

<............... 1 ...................>
- 0 +

Answer 13

aww I hate imaginary numbers. LOL. But okay, I'll bear with it. THANK YOU SO MUCH!..I kinda know how it looks like when it's graphed because I have this graphing calc software, but the most important thing to me right now is why it looked like that.

Okay, I can do this. :))