Ian: New page: Using $v=u+at\ \$ gives $\begin{align} & 15=5+0\textrm{.}2\times t \\ & \\ \end{align}$ where we have used the acceleration calculated in part a). This gives <...

Ian — Wed, 07 Apr 2010 10:38:18 GMT

New page: Using <math>v=u+at\ \</math> gives <math>\begin{align} & 15=5+0\textrm{.}2\times t \\ & \\ \end{align}</math> where we have used the acceleration calculated in part a). This gives <...

New page

Using <math>v=u+at\ \</math> gives

<math>\begin{align}
& 15=5+0\textrm{.}2\times t \\
& \\
\end{align}</math>

where we have used the acceleration calculated in part a).

This gives

<math>t=50\ \text{s}</math>

Solution 6.4b - Revision history

Ian: New page: Using v=u+at\ \ gives \begin{align} & 15=5+0\textrm{.}2\times t \\ & \\ \end{align} where we have used the acceleration calculated in part a). This gives <...

Ian: New page: Using $v=u+at\ \$ gives $\begin{align} & 15=5+0\textrm{.}2\times t \\ & \\ \end{align}$ where we have used the acceleration calculated in part a). This gives <...