Solution 6.4b

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(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 <...)
Current revision (10:38, 7 April 2010) (edit) (undo)
(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 <...)
 

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Using \displaystyle v=u+at\ \ gives

\displaystyle \begin{align} & 15=5+0\textrm{.}2\times t \\ & \\ \end{align}

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

This gives

\displaystyle t=50\ \text{s}