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- 10:30, 20 April 2010 (hist) (diff) Solution 8.10b
- 10:28, 20 April 2010 (hist) (diff) Solution 8.10b
- 10:16, 20 April 2010 (hist) (diff) Solution 8.10b
- 10:09, 20 April 2010 (hist) (diff) Solution 8.10b (New page: We use the equation <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ </math> where, <math>\mathbf{a}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}...)
- 18:50, 19 April 2010 (hist) (diff) Solution 8.10a (top)
- 18:25, 19 April 2010 (hist) (diff) Solution 8.10a
- 17:42, 19 April 2010 (hist) (diff) Solution 8.10a
- 16:32, 19 April 2010 (hist) (diff) Solution 8.10a
- 16:30, 19 April 2010 (hist) (diff) Solution 8.10a
- 16:28, 19 April 2010 (hist) (diff) Solution 8.10a
- 16:25, 19 April 2010 (hist) (diff) Solution 8.10a (New page: We use, <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> to find an expression for the position of the aeroplane at a time t. Here, <math>{{\mathbf{...)
- 15:44, 19 April 2010 (hist) (diff) 8. Constant acceleration equations in vector form
- 15:40, 19 April 2010 (hist) (diff) Solution 8.9b (top)
- 15:26, 19 April 2010 (hist) (diff) Solution 8.9b (New page: We use <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math> with <math>\mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> <math>\mathbf{a}=-10\mathbf{j}</math> <math...)
- 15:18, 19 April 2010 (hist) (diff) Solution 8.9a (top)
- 15:06, 19 April 2010 (hist) (diff) Solution 8.9a
- 14:53, 19 April 2010 (hist) (diff) Solution 8.8a (New page: Here we use the equation <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math> with <math>\mathbf{u}=<math>4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> and <math>\mathbf{v}...) (top)
- 15:39, 18 April 2010 (hist) (diff) Solution 8.8b (top)
- 15:26, 18 April 2010 (hist) (diff) Solution 8.8b
- 15:23, 18 April 2010 (hist) (diff) Solution 8.8b
- 15:18, 18 April 2010 (hist) (diff) Solution 8.8b (New page: We must find the position of the particle after 30 s. Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with <math>\ \mathbf{a}=\mathbf{i}+2\...)
- 15:09, 18 April 2010 (hist) (diff) Solution 8.9a
- 14:45, 18 April 2010 (hist) (diff) m Solution 8.9a (New page: Here we use the equation <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math> with <math>\mathbf{u}=<math>4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> and <math>\mathbf{v}...)
- 12:28, 18 April 2010 (hist) (diff) Solution 8.7c (top)
- 12:17, 18 April 2010 (hist) (diff) Solution 8.7c (New page: The maximum height is when the <math>\mathbf{j}</math> component of the velocity is zero. Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ </math> with <math>\mathbf{u}=8\mathbf{i}+10\mat...)
- 12:14, 18 April 2010 (hist) (diff) Solution 8.7a (top)
- 11:45, 18 April 2010 (hist) (diff) Solution 8.7b (top)
- 11:42, 18 April 2010 (hist) (diff) Solution 8.7b (New page: We use the expression for the position vector obtained in part a), <math> \mathbf{r}=(8\mathbf{i}+10\mathbf{j})t+\frac{1}{2}(-10\mathbf{j}){{t}^{\ 2}} </math> From part a) the ball hits t...)
- 11:39, 18 April 2010 (hist) (diff) Solution 8.7a
- 11:36, 18 April 2010 (hist) (diff) m Solution 8.7a (New page: The ball hits the ground when the <math>\mathbf{j}</math> component of the position vector is zero. We calculate the position vector using <math> \mathbf{r}=\mathbf{u}t+\frac{1}{2}\math...)
- 19:42, 15 April 2010 (hist) (diff) Solution 8.6b
- 18:36, 15 April 2010 (hist) (diff) Solution 8.6b
- 18:30, 15 April 2010 (hist) (diff) Solution 8.6b
- 18:19, 15 April 2010 (hist) (diff) Solution 8.6b
- 18:15, 15 April 2010 (hist) (diff) Solution 8.6b
- 19:57, 14 April 2010 (hist) (diff) Solution 8.6b
- 19:54, 14 April 2010 (hist) (diff) Solution 8.6b
- 19:35, 14 April 2010 (hist) (diff) Solution 8.6b
- 19:31, 14 April 2010 (hist) (diff) Solution 8.6b (New page: Here we use <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math> According to the text <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> We assume th...)
- 19:04, 14 April 2010 (hist) (diff) Solution 8.6a (top)
- 19:02, 14 April 2010 (hist) (diff) Solution 8.6a (New page: <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> and <math>\mathbf{v}= 6\mathbf{i}-8\mathbf{j}</math> when <math>t=10</math>. Also <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math> gives, <...)
- 15:20, 14 April 2010 (hist) (diff) Solution 8.5b (top)
- 15:13, 14 April 2010 (hist) (diff) Solution 8.5b (New page: From part a) we have <math>\mathbf{v}=(80+t\ )\mathbf{i}+4t\ \mathbf{j}</math>. and <math>\mathbf{r}=(80t+\frac{1}{2}{t}^{\ 2}\ )\mathbf{i}+2{t}^{\ 2}\ \mathbf{j}</math> We see the ...)
- 15:02, 14 April 2010 (hist) (diff) Solution 8.5a (top)
- 14:47, 14 April 2010 (hist) (diff) Solution 8.5a
- 14:18, 14 April 2010 (hist) (diff) Solution 8.4b (top)
- 14:14, 14 April 2010 (hist) (diff) Solution 8.5a (New page: We assume the origin is att the point of tacke off which means <math>\ \ \mathbf{ r}_{0}=0</math>. Also we have <math>\mathbf{a}=\mathbf{i}+4\mathbf{j},\ \</math> <math>\mathbf{u}=80\ma...)
- 10:38, 14 April 2010 (hist) (diff) Answer 8.5a (top)
- 10:35, 14 April 2010 (hist) (diff) 8. Exercises (top)
- 09:48, 14 April 2010 (hist) (diff) Answer 7.1b (New page: Image:7.1b.gif) (top)
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