Ian: New page: Using $s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \$ gives $s=0+\frac{1}{2}0\textrm{.}5\times {{10}^{2}}=25\ \text{m}$ Using $v=u+at\ \$ gives $v=0+0\textrm{.}...$

Ian — Thu, 08 Apr 2010 15:43:06 GMT

New page: Using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \</math> gives <math>s=0+\frac{1}{2}0\textrm{.}5\times {{10}^{2}}=25\ \text{m}</math> Using <math>v=u+at\ \</math> gives <math>v=0+0\textrm{.}...

New page

Using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \</math> gives

<math>s=0+\frac{1}{2}0\textrm{.}5\times {{10}^{2}}=25\ \text{m}</math>

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

<math>v=0+0\textrm{.}5\times 10=5\ \text{m}{{\text{s}}^{-1}}</math>

Solution 6.5a - Revision history

Ian: New page: Using s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \ gives s=0+\frac{1}{2}0\textrm{.}5\times {{10}^{2}}=25\ \text{m} Using v=u+at\ \ gives v=0+0\textrm{.}...

Ian: New page: Using $s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \$ gives $s=0+\frac{1}{2}0\textrm{.}5\times {{10}^{2}}=25\ \text{m}$ Using $v=u+at\ \$ gives $v=0+0\textrm{.}...$