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...)
Current revision (18:49, 11 May 2010) (edit) (undo)
 
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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
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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>.
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<math>s=\frac{1}{2}\left( 0 \textrm{.}6 \right)\times 5=1 \textrm{.}5\ \text{m}</math>
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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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use <math>s=\frac{1}{2}(u+v)t</math>. This gives the distance travelled during this stage is
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<math>s=\frac{1}{2}\left(0+ 0 \textrm{.}75 \right)\times 5=1 \textrm{.}875\ \text{m}</math>
The total distance travelled using part b) is
The total distance travelled using part b) is
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<math>37.5+1 \textrm{.}5=39\ \text{m}</math>
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<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}