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Solution 18.1c

From Mechanics

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(New page: Using the result of part b) with <math>a=0</math> gives <math>\begin{align} & 2t-\frac{3{{t}^{2}}}{20}=0 \\ & t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13.3\ \text{s} \\ \end{align}<...)
Line 3: Line 3:
<math>\begin{align}
<math>\begin{align}
& 2t-\frac{3{{t}^{2}}}{20}=0 \\
& 2t-\frac{3{{t}^{2}}}{20}=0 \\
-
& t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13.3\ \text{s} \\
+
& \text{which gives } \\
 +
& t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13\textrm{.}3\ \text{s} \\
\end{align}</math>
\end{align}</math>

Revision as of 15:30, 4 October 2010

Using the result of part b) with a=0 gives

2t−203t2=0which gives t=0 s  or  t=340=13.3 s

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