From Mechanics

Revision as of 15:18, 10 April 2010 (edit) Ian (Talk | contribs) (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...) ← Previous diff		Revision as of 18:07, 11 May 2010 (edit) (undo) Ian (Talk | contribs) Next diff →
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>

---

Revision as of 18:07, 11 May 2010

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 0.075 \text{ m}{{\text{s}}^{-2}}, giving the maximum sped reached is \displaystyle 0+0.075\times 10=0.75\ \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 3.75+1 \textrm{.}875=5.625\ \text{m}