The position of a particle P on a line l is given by s(t) = 2t3 − 4t2 + 2t − 1, where s(t) is in feet and t is in seconds. Find the velocity as a function of time.

Question

v(t)= 6t^2-8t+2

Determine the time intervals (in seconds) in which the particle moves in the positive direction and the negative direction. (Enter your answers as comma-separated lists of intervals.)

Answer 1

s(t) = 2t^3 -4t^2 +2t-1

v(t)=s'(t)= 6t^2-8t+2 >0

3t^2-4t+1>0

(3t-1)(t-1) > 0

v(t)>0 when both factors are the same sign, either both >0 or both<0

t>1/3, & t>1 or t<1/3, & t<1

t>1 or t<1/3

partical moves in a positive direction in the intervals (-infinity,1/3) U (1, infinity)

it moves in a negative direction in the interval (1/3, 1) 1/3<t<1