Solution 19.5a
From Mechanics
(Difference between revisions)
(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
\displaystyle \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}^{2}}}{2} \\ & \\ & \text{ Using}\quad t=40,\ v=5\quad \text{gives} \\ & \\ & 5=20-\frac{k\times {{40}^{2}}}{5} \\ & \\ & 320k=15 \\ & \\ & k=\frac{15}{320}=\frac{3}{64} \end{align}
Note that substituting for \displaystyle k and \displaystyle c in the expression for \displaystyle v we get
\displaystyle v=20-\frac{3{{t}^{2}}}{128}