# 50c.1.PNG - S(t}=2t3—rt2 4t 1 m = 5:2 —14t 4...

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Unformatted text preview: S(t}=2t3—rt2+4t+1 m} = 5:2 —14t+4 arn=12t—14 A{1]=12{1]—14 ar1;=12-14 ar1}——2 Find the ﬁrst and second derivative of the motion equation given. The ﬁrst derivative is the veiooitv'e equation, whiie the second derivative is the aooeieration equation. evn=vrn and em=vrn=arn This is the ﬁrst derivative of the motion equation given and the veiooitv'e eqution. To ﬁnd the acceleration equation calculate the derivative of velocity. This is the acceieraticln equation. To ﬁnd the aooeieration after 15 look at the graph of this eqation, out to ﬁnd the aooeieration at one second, which I think it's what the book really ask for, ﬁnd an}. The ﬂCCEi-Eiﬂti-DI'I at 15 i5 —2I'I"I.Ir5 ...
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