From Mechanics

Revision as of 15:57, 8 April 2010 (edit) Ian (Talk | contribs) (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}}\ \<...) ← Previous diff

Current revision (15:57, 8 April 2010) (edit) (undo) Ian (Talk | contribs) (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}}\ \<...)

---

Current revision

Using \displaystyle v=u+at\ \ gives

\displaystyle v=5+0\textrm{.}2\times 5=6\text{ m}{{\text{s}}^{-1}}

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

\displaystyle s1=5\times 5+\frac{1}{2}0\textrm{.}2\times {{5}^{2}}=27\textrm{.}5\ \text{m}

Together with the first stage the total distance travelled is

\displaystyle 25+27\textrm{.}5=52\textrm{.}5\ \text{m}