Question #4748

4) The position function s of a particle moving along a coordinate line is given by
s(t)=4t^3 +15t^2-18t+1
, where time t is measured in minutes, and
s(t) is
measured in feet. Assume that
the positive direction extends to the right, and the negative direction extends to the
left.
a) Determine the velocity function [rule] v(t) .
b) Determine the time intervals in which the particle moves to the right.
Use a sign chart.
c) Determine the time intervals in which the particle moves to the left.
Use a sign chart.
d) Draw a schematic representing the motion of the particle in the time interval
(-6, 3)

Expert's answer

a)v=ds/dt=12t^2+30t-18

b) particle moves to the right when v>0

12t^2+30t-18>0

12t^2+30t-18=0 when t=(-15+21)/12=1/2 or t=(-15-21)/12=-3

Hence we can say that 12t^2+30t-18>0 when t.1/2 ot t<-3

If we suppose t>0 (physically appropriate assumption) we have only one period of time t>0.5 when v>0 and hence particle moves to the right

when 0<t<0.5 particle moves to the left

d) if -6<t<3 particle moves to the left when -3<t<.5 and to the right all the rest time of the interval

b) particle moves to the right when v>0

12t^2+30t-18>0

12t^2+30t-18=0 when t=(-15+21)/12=1/2 or t=(-15-21)/12=-3

Hence we can say that 12t^2+30t-18>0 when t.1/2 ot t<-3

If we suppose t>0 (physically appropriate assumption) we have only one period of time t>0.5 when v>0 and hence particle moves to the right

when 0<t<0.5 particle moves to the left

d) if -6<t<3 particle moves to the left when -3<t<.5 and to the right all the rest time of the interval

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