Solution 6.5a
From Mechanics
(Difference between revisions)
(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{.}...) |
Current revision
Using \displaystyle s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \ gives
\displaystyle s=0+\frac{1}{2}0\textrm{.}5\times {{10}^{2}}=25\ \text{m}
Using \displaystyle v=u+at\ \ gives
\displaystyle v=0+0\textrm{.}5\times 10=5\ \text{m}{{\text{s}}^{-1}}