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Solution 19.5a

From Mechanics

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(New page: <math>\begin{align} & v=\int{adt} \\ & \\ & =\int{\left( -kt \right)dt} \\ & \\ & =-\frac{k{{t}^{2}}}{2}+c \\ & \\ & t=0,\ v=20\ \Rightarrow \ c=20 \\ & \\ & v=20-\frac{k{{t}^{...)
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& k=\frac{15}{320}=\frac{3}{64}
& k=\frac{15}{320}=\frac{3}{64}
\end{align}</math>
\end{align}</math>
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Note that substituting for <math>k</math> and <math>c</math> in the expression for <math>v</math> we get
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<math>v=20-\frac{3{{t}^{2}}}{128}</math>

Revision as of 15:43, 11 October 2010

v=adt=−ktdt=−2kt2+ct=0 v=20  c=20v=20−2kt2 Usingt=40 v=5gives5=20−5k402320k=15k=15320=364

Note that substituting for k and c in the expression for v we get

v=20−3t2128

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