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anonymous
 5 years ago
How do I solve for #12? I've attached the problem.
anonymous
 5 years ago
How do I solve for #12? I've attached the problem.

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0t = 8 is the maximum acceleration

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha thanks do you know how to do #12?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0acceleration is any change in velocity; where the velocity is level, there is no acceleration.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0the answer to 12 is D if that helps

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think your looking at #10

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol....without knowing what D stands for, it aint much help ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0oh.... theres more if you scroll it lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0at t=8 the objects position was at x=10 ?? how do they figure that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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<t<8 now we need position function x(t) so integrate v(t) x(t) = t^2+12t +c we know x(8) = 10 8^2 +12*8 +c = 10 c = 22 find x(5) 5^2 +12*5 22 = 13
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