Question 1081862
a)
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*[illustration gg3.JPG].
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Looking at the graph, the particle travels from 9 to 7 in the 1st second, from 7 to 9 in the 2nd second and from 9 to 15 in the 3rd second.
So it traveled,
{{{D=abs(7-9)+abs(9-7)+abs(15-9)=2+2+6=10}}}{{{m}}}
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b) Find the instantaneous velocity when {{{t=3}}},
{{{v=ds/dt=4t-4=4(3)-4=12-4=8}}}{{{m/s}}}
So then integrate the new acceleration to get the new velocity,
{{{v=int(a,dt,t=3,t)=int((t-8),dt,t=3,t)=t^2/2-8t+C}}}
When {{{t=3}}},{{{v=8}}},
{{{8=(3)^2/2-8(3)+C}}}
{{{8=9/2-24+C}}}
{{{C=32-9/2}}}
{{{C=64/2-9/2}}}
{{{C=55/2}}}
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{{{v(t)=t^2/2-8t+55/2}}} for {{{t>=3}}}
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Graphing this and looking for the next crossing of the x-axis,
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*[illustration gg4.JPG].
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{{{t=5}}}{{{s}}}

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You could also factor,
{{{t^2/2-8t+55/2=0}}}
{{{t^2-16t+55=0}}}
{{{(t-5)(t-11)=0}}}
{{{t=5}}} and {{{t=11}}}