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- 09:16, 11 October 2010 (hist) (diff) Solution 19.2a (New page: <math>\begin{align} & v=\int{a}dt=\frac{{{t}^{2}}}{50}+c \\ & \\ & t=0, \ \ v=2\ \Rightarrow \ c=2 \\ & \\ & v=\frac{{{t}^{2}}}{50}+2 \\ \end{align}</math>) (top)
- 09:14, 11 October 2010 (hist) (diff) Answer 19.2b (New page: 933 m) (top)
- 09:13, 11 October 2010 (hist) (diff) Answer 19.2a (New page: <math>\frac{{{t}^{2}}}{50}+2</math>) (top)
- 17:46, 10 October 2010 (hist) (diff) Solution 19.1c (New page: <math>\begin{align} & s(15)=\int_{0}^{15}{v}dt \\ & \\ & =\int_{0}^{15}{( t-\frac{{{t}^{2}}}{30} )dt} \\ & \\ & =\left[ \frac{{{t}^{2}}}{2}-\frac{{{t}^{3}}}{90} \right]_{0}^{15} \\ & ...)
- 17:42, 10 October 2010 (hist) (diff) Solution 19.1b (New page: <math>v(15)=15-\frac{{{15}^{2}}}{30}=7\textrm{.}5\text{ m}{{\text{s}}^{\text{-1}}}</math>) (top)
- 17:40, 10 October 2010 (hist) (diff) Solution 19.1a (New page: <math>\begin{align} & v=\int{a}dt=t-\frac{{{t}^{2}}}{30}+c \\ & t=0,\ \ v=0\Rightarrow c=0 \\ & v=t-\frac{{{t}^{2}}}{30} \\ \end{align}</math>) (top)
- 17:38, 10 October 2010 (hist) (diff) Answer 19.1c (New page: 75 m) (top)
- 17:38, 10 October 2010 (hist) (diff) Answer 19.1b (New page: <math>\text{7.}\text{5 m}{{\text{s}}^{-\text{1}}}</math>) (top)
- 17:36, 10 October 2010 (hist) (diff) Answer 19.1a (New page: <math>t-\frac{{{t}^{ 2}}}{30}</math>) (top)
- 17:26, 10 October 2010 (hist) (diff) 18. Motion with variable acceleration I
- 17:20, 10 October 2010 (hist) (diff) Solution 18.9c (New page: <math>\begin{align} & \mathbf{a}=\frac{d\mathbf{v}}{dt}=2\mathbf{i}+(6t-24)\mathbf{j} \\ & \mathbf{F}=3\mathbf{a}=6\mathbf{i}+(18t-72)\mathbf{j} \\ \end{align}</math>) (top)
- 17:18, 10 October 2010 (hist) (diff) Solution 18.9b (New page: <math>\begin{align} & 3{{t}^{2}}-24t+45=0 \\ & {{t}^{2}}-8t+15=0 \\ & (t-3)(t-5)=0 \\ & t=3\ \text{s or }t=5\ \text{s} \\ \end{align}</math>) (top)
- 17:16, 10 October 2010 (hist) (diff) Solution 18.9a (New page: <math>\mathbf{v}=\frac{d\mathbf{r}}{dt}=(2t-1)\mathbf{i}+(3{{t}^{2}}-24t+45)\mathbf{j}</math>) (top)
- 17:16, 10 October 2010 (hist) (diff) Answer 18.9c (New page: <math>6\mathbf{i}+(18t-72)\mathbf{j}</math>) (top)
- 17:14, 10 October 2010 (hist) (diff) Answer 18.9b (New page: <math>t=3\ \text{s or }t=5\ \text{s}</math>) (top)
- 17:12, 10 October 2010 (hist) (diff) Answer 18.9a (New page: <math>(2t-1)\mathbf{i}+(3{{t}^{2}}-24t+45)\mathbf{j}</math>) (top)
- 18:52, 8 October 2010 (hist) (diff) Solution 18.8d (New page: <math>\mathbf{a}=\frac{d\mathbf{v}}{dt}=-\frac{1}{5}\mathbf{i}\ \text{m}{{\text{s}}^{-2}}</math> This is a constant vector pointing in the opposite direction to <math>\mathbf{i}</math>, ...) (top)
- 18:49, 8 October 2010 (hist) (diff) Solution 18.8c (New page: From part b) we have <math>\mathbf{v}=\left( 2-\frac{t}{5} \right)\mathbf{i}+2\mathbf{j}</math> If the boat is travelling north, the <math>\mathbf{i}</math> component of the velocity mu...) (top)
- 18:42, 8 October 2010 (hist) (diff) Solution 18.8b (New page: <math>\begin{align} & \mathbf{v}=\frac{d\mathbf{r}}{dt}=\left( 2-\frac{t}{5} \right)\mathbf{i}+2\mathbf{j} \\ & v=\sqrt{{{\left( 2-\frac{t}{5} \right)}^{2}}+{{2}^{2}}}=\sqrt{4-\frac{4t}{5...)
- 18:39, 8 October 2010 (hist) (diff) Solution 18.8a (New page: <math>\begin{align} & 2t-\frac{{{t}^{2}}}{10}=0 \\ & t\left( 2-\frac{t}{10} \right)=0 \\ & t=0\text{ or }t=20 \\ \end{align}</math> <math>t=0</math> is not of physical interest thus th...) (top)
- 18:35, 8 October 2010 (hist) (diff) Answer 18.8d (New page: Due West) (top)
- 18:35, 8 October 2010 (hist) (diff) Answer 18.8c (New page: 10 s) (top)
- 18:34, 8 October 2010 (hist) (diff) Answer 18.8b (New page: <math>\sqrt{4-\frac{4t}{5}+\frac{{{t}^{2}}}{25}}</math>)
- 18:33, 8 October 2010 (hist) (diff) Answer 18.8a (New page: 20 s) (top)
- 18:31, 8 October 2010 (hist) (diff) Solution 18.7c (New page: <math>\mathbf{a}=\frac{d\mathbf{v}}{dt}=6\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math> This is a constant vector with magnitude <math>6\ \text{m}{{\text{s}}^{-2}}</math> and points vertica...) (top)
- 18:23, 8 October 2010 (hist) (diff) Solution 18.7b (New page: <math>\begin{align} & \mathbf{v}=\frac{d\mathbf{r}}{dt}=40\mathbf{i}+(6t+20)\mathbf{j} \\ & \mathbf{v}(10)=40\mathbf{i}+(6\times 10+20)\mathbf{j}=40\mathbf{i}+80\mathbf{j} \\ & v(10)=\sq...) (top)
- 18:21, 8 October 2010 (hist) (diff) Solution 18.7a (New page: <math>\begin{align} & 3{{t}^{2}}+20t=500 \\ & 3{{t}^{2}}+20t-500=0 \\ & (3t+50)(t-10)=0 \\ & t=-\frac{50}{3}\text{ or }t=10 \\ & \text{It is at a height of 500 m when }t=10 \text { s...) (top)
- 18:14, 8 October 2010 (hist) (diff) Answer 18.7c (New page: <math>6\ \text{m}{{\text{s}}^{-2}}</math>) (top)
- 18:11, 8 October 2010 (hist) (diff) Answer 18.7b (New page: <math>\text{89}\textrm{.}\text{4 m}{{\text{s}}^{-\text{1}}}</math>) (top)
- 18:07, 8 October 2010 (hist) (diff) Answer 18.7a (New page: 10 s) (top)
- 18:05, 8 October 2010 (hist) (diff) Solution 18.6c (top)
- 20:03, 7 October 2010 (hist) (diff) 18. Exercises
- 20:01, 7 October 2010 (hist) (diff) 18. Exercises
- 19:48, 7 October 2010 (hist) (diff) Answer 18.6a (top)
- 19:47, 7 October 2010 (hist) (diff) Solution 18.6c (New page: <math>\begin{align} & \mathbf{a}=\frac{d\mathbf{v}}{dt}=-4\mathbf{i}-10\mathbf{j} \\ & \mathbf{F}=5(-4\mathbf{i}-10\mathbf{j})=-20\mathbf{i}-50\mathbf{j} \\ & F=\sqrt{{{20}^{2}}+{{50}^{2...)
- 19:45, 7 October 2010 (hist) (diff) Solution 18.6b (New page: <math>\begin{align} & \mathbf{v}=\frac{d\mathbf{r}}{dt}=(-4t)\mathbf{i}+(-10t)\mathbf{j} \\ & \mathbf{v}(2)=-8\mathbf{i}-20\mathbf{j} \\ & v(2)=\sqrt{{{8}^{2}}+{{20}^{2}}}=21\textrm{.}5\...) (top)
- 19:44, 7 October 2010 (hist) (diff) Solution 18.6a (New page: <math>\begin{align} & \mathbf{r}(10)=(12-2\times {{10}^{2}})\mathbf{i}+(10-5\times {{10}^{2}})\mathbf{j} \\ & =(-188\mathbf{i}-490\mathbf{j})\ \text{m} \end{align}</math>) (top)
- 19:41, 7 October 2010 (hist) (diff) Answer 18.6c (New page: 53.9 N) (top)
- 19:41, 7 October 2010 (hist) (diff) Answer 18.6b (New page: <math>\text{21}\textrm{.}\text{5 m}{{\text{s}}^{-\text{1}}}</math>) (top)
- 19:37, 7 October 2010 (hist) (diff) Answer 18.6a (New page: <math>\left( -\text{188}\mathbf{i}-\text{ 49}0\mathbf{j} \right)\ \text{m}</math>)
- 18:57, 7 October 2010 (hist) (diff) 18. Exercises
- 18:34, 6 October 2010 (hist) (diff) Solution 18.5d (New page: Using <math>a=100-10t</math> from part c) when <math>a=0</math> then <math>t=10\ \text{s}</math>. Alternatively this can be read off from the sketch in part c.) (top)
- 18:30, 6 October 2010 (hist) (diff) Answer 18.5d (New page: 10 s) (top)
- 18:28, 6 October 2010 (hist) (diff) Hint 18.5c (New page: <math>\begin{align} & a=\frac{dv}{dt}=2k-10t \\ & =100-10t \end{align}</math> Then plot <math>a</math> against <math>t</math>) (top)
- 18:26, 6 October 2010 (hist) (diff) Image:18.5c.gif (top)
- 18:25, 6 October 2010 (hist) (diff) Answer 18.5c (New page: Image:18.5c.gif) (top)
- 18:23, 6 October 2010 (hist) (diff) Answer 18.5b (New page: <math>k=50</math>) (top)
- 18:22, 6 October 2010 (hist) (diff) Solution 18.5b (New page: From part a) <math>v=2kt-5{{t}^{2}}</math> Thus <math>\begin{align} & 0=2k\times 20-5\times {{20}^{2}} \\ & 0=40k-2000 \\ & k=\frac{2000}{40}=50 \\ \end{align}</math>) (top)
- 18:18, 6 October 2010 (hist) (diff) Solution 18.5a (New page: <math>\begin{align} & v=\frac{ds}{dt}=2kt-5{{t}^{2}} \\ & v(0)=2k\times 0-5\times {{0}^{2}}=0 \\ \end{align}</math>) (top)
- 17:52, 6 October 2010 (hist) (diff) Solution 18.4c (New page: From part a) <math>a=\frac{dv}{dt}=36-6t</math> We see <math>a</math> decreases as the time <math>t</math> increases. Thus the maximum acxceleration is at <math>t=0</math> giving <math...)
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