Solution 6.8c
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(New page: For the second stage where the lift slows down we use <math>s=\frac{1}{2}(u+v)t</math>. This gives the distance travelled during this stage is <math>s=\frac{1}{2}\left( 0 \textrm{.}6 \rig...)
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Revision as of 15:18, 10 April 2010
For the second stage where the lift slows down we use \displaystyle s=\frac{1}{2}(u+v)t. This gives the distance travelled during this stage is
\displaystyle s=\frac{1}{2}\left( 0 \textrm{.}6 \right)\times 5=1 \textrm{.}5\ \text{m}
The total distance travelled using part b) is
\displaystyle 37.5+1 \textrm{.}5=39\ \text{m}
