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- 09:46, 14 April 2010 (hist) (diff) Solution 7.1c (New page: The red spot shows where the ball hits the ground Image:7.1b.gif) (top)
- 09:44, 14 April 2010 (hist) (diff) Image:7.1b.gif (top)
- 16:48, 13 April 2010 (hist) (diff) Solution 8.4c (top)
- 16:38, 13 April 2010 (hist) (diff) Solution 8.4c
- 16:26, 13 April 2010 (hist) (diff) Solution 8.4c (New page: We have <math>\mathbf{a}=4\mathbf{i}– 8\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\ \ \mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}</math>. The ...)
- 16:19, 13 April 2010 (hist) (diff) Solution 8.4b (New page: Here we have <math>\mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\ \ \mathbf{ r}_{0}=6\mathbf{i}...)
- 16:05, 13 April 2010 (hist) (diff) Solution 8.4a
- 15:57, 13 April 2010 (hist) (diff) Solution 8.4a
- 15:48, 13 April 2010 (hist) (diff) Solution 8.4a
- 15:43, 13 April 2010 (hist) (diff) Solution 8.4a
- 15:42, 13 April 2010 (hist) (diff) Solution 8.4a (New page: Here we have <math>\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>{\mathbf{ r}}_{0}=6\mathbf{i}+2\mathbf{j}</math>.)
- 15:28, 13 April 2010 (hist) (diff) Solution 8.1b (top)
- 15:27, 13 April 2010 (hist) (diff) Solution 8.1a
- 15:26, 13 April 2010 (hist) (diff) Solution 8.1a
- 15:26, 13 April 2010 (hist) (diff) Solution 8.1b
- 15:25, 13 April 2010 (hist) (diff) Solution 8.3 (top)
- 15:20, 13 April 2010 (hist) (diff) Solution 8.3 (New page: In this case we have <math>\mathbf{u}=0</math> , <math>\mathbf{a}=2\mathbf{i}+3\mathbf{j}</math> and <math>{{\mathbf{r}}_{0}}=0</math>. Substituting these into the equation <math>\math...)
- 15:05, 13 April 2010 (hist) (diff) Solution 8.2b (New page: In this case we have <math>\mathbf{u}=5\mathbf{i}</math> , <math>\mathbf{a}=-4\mathbf{j}</math> and <math>{{\mathbf{r}}_{0}}=18\mathbf{i}+14\mathbf{j}</math>. Substituting these into ...) (top)
- 14:58, 13 April 2010 (hist) (diff) Solution 8.2a (New page: In this case we have <math>\mathbf{u}=5\mathbf{i}</math> and <math>\mathbf{a}=-4\mathbf{j}</math> Substituting these into the equation <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ \</math> ...) (top)
- 14:50, 13 April 2010 (hist) (diff) Solution 8.1b
- 13:56, 13 April 2010 (hist) (diff) 7. Exercises
- 17:44, 12 April 2010 (hist) (diff) Solution 8.1a
- 17:24, 12 April 2010 (hist) (diff) Solution 8.1a (Removing all content from page)
- 17:24, 12 April 2010 (hist) (diff) Solution 8.1b (New page: In this case we have <math>\mathbf{u}=4\mathbf{i}</math> , <math>\mathbf{a}=0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j}</math> and <math>{{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}...)
- 17:20, 12 April 2010 (hist) (diff) Solution 8.1a
- 17:15, 12 April 2010 (hist) (diff) Solution 8.1a
- 17:10, 12 April 2010 (hist) (diff) Solution 8.1a (New page: In this case we have <math>\mathbf{u}=4\mathbf{i}</math> , <math>\mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j}</math> and <math>{{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}</math>. Substit...)
- 16:50, 12 April 2010 (hist) (diff) 8. Constant acceleration equations in vector form
- 16:37, 12 April 2010 (hist) (diff) 8. Constant acceleration equations in vector form
- 16:24, 11 April 2010 (hist) (diff) Solution 7.4c (New page: We insert <math>t=0\ \text{s}</math> in the expression for the position vector to find the point where the frisby waqs thrown from. <math>\begin{align} & \mathbf{r}=\left( 5\times 0-{{0...)
- 16:21, 11 April 2010 (hist) (diff) Solution 7.4b (New page: We insert <math>t=8\ \text{s}</math> in the expression for the position vector, <math>\begin{align} & \mathbf{r}=\left( 5\times 8-{{8}^{2}} \right)\mathbf{i}+\left( 7\times 8 \right)\mat...) (top)
- 16:15, 11 April 2010 (hist) (diff) Solution 7.4a (New page: The frisby hits the ground when the <math>\mathbf{k}</math> term is zero. This gives <math>\begin{align} & 24+5t-{{t}^{\ 2}}=0 \\ & \\ & {{t}^{\ 2}}-5-24=0 \\ \end{align}</math> ...)
- 13:31, 11 April 2010 (hist) (diff) Solution 7.3d (New page: If the boat is due east of the rock this means the boat must have the same <math>\mathbf{j}</math> term as the rock. <math>4t-12=352</math> giving <math>t=91\ \text{s}</math>) (top)
- 13:21, 11 April 2010 (hist) (diff) Solution 7.3c (top)
- 12:43, 11 April 2010 (hist) (diff) Solution 7.3c (New page: If the boat is due north of the rock this means the boat must have the same <math>\mathbf{i}</math> coordinate as the rock. <math>2t-16=174</math> giving <math>t=95\ \text{s}</math...)
- 12:17, 11 April 2010 (hist) (diff) 7. Position Vectors (top)
- 11:25, 11 April 2010 (hist) (diff) Solution 7.3a (New page: <math>t=0\ \</math> gives <math>\mathbf{r}=\left( 2\times 0-16 \right)\mathbf{i}+\left( 4\times 0-12 \right)\mathbf{j}=-16\mathbf{i}-12\mathbf{j}\ \text{m}</math> <math>t=40\ \</math> g...) (top)
- 10:29, 11 April 2010 (hist) (diff) Solution 7.1a (top)
- 10:24, 11 April 2010 (hist) (diff) Solution 7.1a (New page: <math>t=0\ \</math> gives <math>\mathbf{r}=6\times 0\mathbf{i}+\left( 15\times 0-4.9\times {{0}^{2}} \right)\mathbf{j}=0\mathbf{i}+0\mathbf{j}\ \text{m}</math> <math>t=1\ \</math> gives...)
- 15:46, 10 April 2010 (hist) (diff) 6. Exercises
- 15:18, 10 April 2010 (hist) (diff) Solution 6.8c (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...)
- 15:09, 10 April 2010 (hist) (diff) Solution 6.8b
- 13:09, 10 April 2010 (hist) (diff) m 2. Introduction to force and gravity
- 12:54, 10 April 2010 (hist) (diff) Solution 6.8b (New page: We investigate the distance after <math>10\ \text{s}</math>.Using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> gives <math>s=0+\frac{1}{2}\times \left( 0.075 \right)\times {{10}^{2}}</math>)
- 12:42, 10 April 2010 (hist) (diff) Solution 6.8a (top)
- 12:38, 10 April 2010 (hist) (diff) Solution 6.10c (top)
- 12:30, 10 April 2010 (hist) (diff) Solution 6.10c (New page: We use the results of the first two parts of the question. As the sprinter runs <math>\text{16 m}</math> in the first stage of the race he must run <math>\text{84 m}</math> in the secon...)
- 12:24, 10 April 2010 (hist) (diff) Solution 6.10b (New page: Using <math>v=u+at</math> gives <math>v=0+2\times 4=8\ \text{m}{{\text{s}}^{-1}}</math>) (top)
- 12:22, 10 April 2010 (hist) (diff) Solution 6.10a (New page: Using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> gives <math>s=0+\frac{1}{2}\times 2\times {{4}^{2}}=16\ \text{m}</math>) (top)
- 12:12, 10 April 2010 (hist) (diff) Solution 6.9 (New page: Using the equation <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}</math> gives <math>\begin{align} & 162=36t+\frac{1}{2}\times \left( -3 \right){{t}^{\ 2}} \\ & \\ & 324=72t-3{{t}^{\ 2}} \\ & \...) (top)
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