Unformatted text preview: S(t}=2t3—rt2+4t+1 m} = 5:2 —14t+4 arn=12t—14 A{1]=12{1]—14
ar1;=1214
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 ﬂCCEiEiﬂtiDI'I at 15 i5 —2I'I"I.Ir5 ...
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 Spring '16
 Epi
 Calculus, Chapter 3, Section 1, Early Transcendentals

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