Ian: New page: Using $v=u+at\ \$ gives $v=5+0\textrm{.}2\times 5=6\text{ m}{{\text{s}}^{-1}}$ The distance travelled during this stage using $s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \<...$

Ian — Thu, 08 Apr 2010 15:57:19 GMT

New page: Using <math>v=u+at\ \</math> gives <math>v=5+0\textrm{.}2\times 5=6\text{ m}{{\text{s}}^{-1}}</math> The distance travelled during this stage using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \<...

New page

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

<math>v=5+0\textrm{.}2\times 5=6\text{ m}{{\text{s}}^{-1}}</math>

The distance travelled during this stage using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \</math> is

<math>s1=5\times 5+\frac{1}{2}0\textrm{.}2\times {{5}^{2}}=27\textrm{.}5\ \text{m}</math>

Together with the first stage the total distance travelled is

<math>25+27\textrm{.}5=52\textrm{.}5\ \text{m}</math>

Solution 6.5b - Revision history

Ian: New page: Using v=u+at\ \ gives v=5+0\textrm{.}2\times 5=6\text{ m}{{\text{s}}^{-1}} The distance travelled during this stage using s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \<...

Ian: New page: Using $v=u+at\ \$ gives $v=5+0\textrm{.}2\times 5=6\text{ m}{{\text{s}}^{-1}}$ The distance travelled during this stage using $s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \<...$