Solution 6.8c - Revision history

Ian at 18:49, 11 May 2010

Ian — Tue, 11 May 2010 18:49:06 GMT

←Older revision		Revision as of 18:49, 11 May 2010
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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>

Ian at 18:07, 11 May 2010

Ian — Tue, 11 May 2010 18:07:52 GMT

←Older revision		Revision as of 18:07, 11 May 2010
Line 1:		Line 1:
-	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~~	+	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>.

-	<math>s=\frac{1}{2}\left( 0 \textrm{.}6 \right)\times 5=1 \textrm{.}5\ \text{m}</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>
		+
		+
		+	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+ 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

-	<math>37.5+1 \textrm{.}5=39\ \text{m}</math>	+	<math>3.75+1 \textrm{.}875=5.625\ \text{m}</math>

Ian: New page: For the second stage where the lift slows down we use $s=\frac{1}{2}(u+v)t$ . This gives the distance travelled during this stage is $s=\frac{1}{2}\left( 0 \textrm{.}6 \rig...$

Ian — Sat, 10 Apr 2010 15:18:10 GMT

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...

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 \right)\times 5=1 \textrm{.}5\ \text{m}</math>

The total distance travelled using part b) is

<math>37.5+1 \textrm{.}5=39\ \text{m}</math>

Solution 6.8c - Revision history

Ian at 18:49, 11 May 2010

Ian at 18:07, 11 May 2010

Ian: New page: For the second stage where the lift slows down we use s=\frac{1}{2}(u+v)t. This gives the distance travelled during this stage is s=\frac{1}{2}\left( 0 \textrm{.}6 \rig...

Ian: New page: For the second stage where the lift slows down we use $s=\frac{1}{2}(u+v)t$ . This gives the distance travelled during this stage is $s=\frac{1}{2}\left( 0 \textrm{.}6 \rig...$