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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}^{...)
Current revision (17:29, 27 March 2011) (edit) (undo)
 
(One intermediate revision not shown.)
Line 14: Line 14:
& 5=20-\frac{k\times {{40}^{2}}}{5} \\
& 5=20-\frac{k\times {{40}^{2}}}{5} \\
& \\
& \\
-
& 320k=15 \\
+
& 800k=15 \\
& \\
& \\
-
& k=\frac{15}{320}=\frac{3}{64}
+
& k=\frac{15}{800}=\frac{3}{160}
\end{align}</math>
\end{align}</math>
 +
 +
Note that substituting for <math>k</math> and <math>c</math> in the expression for <math>v</math> we get
 +
 +
<math>v=20-\frac{3{{t}^{2}}}{320}</math>

Current revision

v=adt=−ktdt=−2kt2+ct=0 v=20  c=20v=20−2kt2 Usingt=40 v=5gives5=20−5k402800k=15k=15800=3160

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

v=20−3t2320

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