anonymous
  • anonymous
How do I solve for #12? I've attached the problem.
Mathematics
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anonymous
  • anonymous
How do I solve for #12? I've attached the problem.
Mathematics
katieb
  • katieb
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
1 Attachment
amistre64
  • amistre64
t = 8 is the maximum acceleration
amistre64
  • amistre64
maybe :)

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anonymous
  • anonymous
haha thanks do you know how to do #12?
amistre64
  • amistre64
acceleration is any change in velocity; where the velocity is level, there is no acceleration.
amistre64
  • amistre64
we have the velocity changing at a rate of 1 in the first part. zero in the next segment; 6/3 in the next = 2 and 4/1 in the last part. so maybe between 8 and 9
anonymous
  • anonymous
the answer to 12 is D if that helps
anonymous
  • anonymous
i think your looking at #10
amistre64
  • amistre64
lol....without knowing what D stands for, it aint much help ;)
amistre64
  • amistre64
oh.... theres more if you scroll it lol
amistre64
  • amistre64
at t=8 the objects position was at x=10 ?? how do they figure that?
dumbcow
  • dumbcow
ok for #12 find equation for line from t=5 and t=8 slope is -2 v= 2 when t=5 2=-2*5 + b b = 12 v(t) = -2t +12 for 5

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