Solution 6.8c
From Mechanics
(Difference between revisions)
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For the second stage where the lift slows down we first must calculate the maximum speed which is reached after 10 seconds using <math>v=u+at</math>. | For the second stage where the lift slows down we first must calculate the maximum speed which is reached after 10 seconds using <math>v=u+at</math>. | ||
| - | From part a) we have the acceleration is <math>0.075 \text{ m}{{\text{s}}^{-2}}</math>, giving the maximum sped reached is <math>0+0.075\times 10=0.75\ \text{m}{{\text{s}}^{-1}}</math> | + | From part a) we have the acceleration is <math>0\textrm{.}075 \text{ m}{{\text{s}}^{-2}}</math>, giving the maximum sped reached is <math>0+0\textrm{.}075\times 10=0\textrm{.}75\ \text{m}{{\text{s}}^{-1}}</math> |
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The total distance travelled using part b) is | The total distance travelled using part b) is | ||
| - | <math>3.75+1 \textrm{.}875=5.625\ \text{m}</math> | + | <math>3\textrm{.}75+1 \textrm{.}875=5\textrm{.}625\ \text{m}</math> |
Current revision
For the second stage where the lift slows down we first must calculate the maximum speed which is reached after 10 seconds using \displaystyle v=u+at.
From part a) we have the acceleration is \displaystyle 0\textrm{.}075 \text{ m}{{\text{s}}^{-2}}, giving the maximum sped reached is \displaystyle 0+0\textrm{.}075\times 10=0\textrm{.}75\ \text{m}{{\text{s}}^{-1}}
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+ 0 \textrm{.}75 \right)\times 5=1 \textrm{.}875\ \text{m}
The total distance travelled using part b) is
\displaystyle 3\textrm{.}75+1 \textrm{.}875=5\textrm{.}625\ \text{m}
