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- 13:04, 12 October 2010 (hist) (diff) Solution 20.2 (New page: <math>F=m\frac{{{v}^{2}}}{r}=0\textrm{.}1\times \frac{{{2}^{2}}}{0\textrm{.}2}=2\text{ N}</math>) (top)
- 13:03, 12 October 2010 (hist) (diff) Solution 20.1 (New page: Firstly, the angular speed must be converted to standard units. <math>\omega =400\text{ rpm }=\text{ }\frac{400\times 2\pi }{\text{60}}=\frac{40\pi }{3}\text{ rad}{{\text{s}}^{\text{-1}}}...) (top)
- 07:39, 12 October 2010 (hist) (diff) Solution 19.7a (top)
- 17:13, 11 October 2010 (hist) (diff) Solution 19.8d
- 17:09, 11 October 2010 (hist) (diff) Solution 19.8c (top)
- 17:09, 11 October 2010 (hist) (diff) Solution 19.8d (New page: The expression for the position vector <math>\mathbf{r}</math> has been obtained in part b). <math>\mathbf{r}=\left( 40t \right)\mathbf{i}+\left( -2t+400 \right)\mathbf{j}</math>)
- 17:06, 11 October 2010 (hist) (diff) Solution 19.8c
- 16:55, 11 October 2010 (hist) (diff) Solution 19.8c
- 16:54, 11 October 2010 (hist) (diff) Solution 19.8c (New page: The expression for the position vector<math>\mathbf{r}</math> has been obtained in part b),)
- 16:52, 11 October 2010 (hist) (diff) Solution 19.8b (New page: The expression for the velocity <math>\mathbf{v}</math> was obtained in part a), <math>\begin{align} & \mathbf{r}=\int{\mathbf{v}dt} \\ & \\ & =\left( 40t+{{d}_{1}} \right)\mathbf{i}+\...)
- 16:50, 11 October 2010 (hist) (diff) Solution 19.8a (New page: <math>\begin{align} & \mathbf{v}=\int{\mathbf{a}dt} \\ & \\ & =\left( {{c}_{1}} \right)\mathbf{i}+\left( -2t+{{c}_{2}} \right)\mathbf{j} \\ & \\ & t=0,\ \mathbf{v}=40\mathbf{i}\ \Rig...) (top)
- 16:47, 11 October 2010 (hist) (diff) Answer 19.8d (New page: 200 s)
- 16:47, 11 October 2010 (hist) (diff) Answer 19.8c (New page: 20 s) (top)
- 16:47, 11 October 2010 (hist) (diff) Answer 19.8b (New page: <math>\left( 40t \right)\mathbf{i}+\left( -2t+400 \right)\mathbf{j}</math>)
- 16:45, 11 October 2010 (hist) (diff) Answer 19.8a (New page: <math>\left( 40 \right)\mathbf{i}+\left( -2t \right)\mathbf{j}</math>) (top)
- 16:34, 11 October 2010 (hist) (diff) Solution 19.7d (New page: Using the result for the position vector <math>\mathbf{r}</math> from part c), <math>\begin{align} & \mathbf{r}(20)=\left( \frac{{{20}^{3}}}{2}-8\times 20+90 \right)\mathbf{i}+\left( \fra...) (top)
- 16:31, 11 October 2010 (hist) (diff) Solution 19.7c (New page: The expression for the velocity <math>\mathbf{v}</math> has been obtained in part b), <math>\begin{align} & \mathbf{r}=\int{\mathbf{v}}dt \\ & \\ & =\left( \frac{{{t}^{3}}}{2}-8t+{{d}_...) (top)
- 16:28, 11 October 2010 (hist) (diff) Solution 19.7b (New page: The expression for the acceleration <math>\mathbf{a}</math> has been obtained in part a), <math>\begin{align} & \mathbf{v}=\int{\mathbf{a}dt} \\ & \\ & =\left( \frac{3{{t}^{2}}}{2}+{{c...) (top)
- 16:05, 11 October 2010 (hist) (diff) Solution 19.7a (New page: <math>\mathbf{a}=\frac{\mathbf{F}}{4}=\frac{12t\mathbf{i}+(6-5t)\mathbf{j}}{4}=3t\mathbf{i}+\left( \frac{3}{2}-\frac{5t}{4} \right)\mathbf{j}</math>)
- 16:01, 11 October 2010 (hist) (diff) Answer 19.7d (New page: 4154 m) (top)
- 16:01, 11 October 2010 (hist) (diff) Answer 19.7c (New page: <math>\left( \frac{{{t}^{3}}}{2}-8t+90 \right)\mathbf{i}+\left( \frac{3{{t}^{2}}}{4}-\frac{5{{t}^{3}}}{24}+20 \right)\mathbf{j}</math>) (top)
- 16:00, 11 October 2010 (hist) (diff) Answer 19.7b (New page: <math>\left( \frac{3{{t}^{2}}}{2}-8 \right)\mathbf{i}+\left( \frac{3t}{2}-\frac{5{{t}^{2}}}{8} \right)\mathbf{j}</math>) (top)
- 15:58, 11 October 2010 (hist) (diff) Answer 19.7a (New page: <math>3t\mathbf{i}+\left( \frac{3}{2}-\frac{5t}{4} \right)\mathbf{j}</math>) (top)
- 15:57, 11 October 2010 (hist) (diff) Solution 19.6b (New page: <math>\begin{align} & s=\int_{0}^{10}{\left( 3{{t}^{2}}+2t+5 \right)}dt \\ & \\ & =\left[ {{t}^{3}}+{{t}^{2}}+5t \right]_{0}^{10} \\ & \\ & =\left( {{10}^{3}}+{{10}^{2}}+5\times 10 \...) (top)
- 15:55, 11 October 2010 (hist) (diff) Solution 19.6a (New page: <math>a=\frac{dv}{dt}=6t+2</math>) (top)
- 15:52, 11 October 2010 (hist) (diff) Answer 19.6b (New page: 1150 m) (top)
- 15:51, 11 October 2010 (hist) (diff) Answer 19.6a (New page: <math>6t+2</math>) (top)
- 15:50, 11 October 2010 (hist) (diff) Solution 19.5b (New page: Using the expression for <math>v</math> obtained in part a), <math>\begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{12...)
- 15:43, 11 October 2010 (hist) (diff) Solution 19.5a
- 15:37, 11 October 2010 (hist) (diff) Solution 19.5a (New page: <math>\begin{align} & v=\int{adt} \\ & \\ & =\int{\left( -kt \right)dt} \\ & \\ & =-\frac{k{{t}^{2}}}{2}+c \\ & \\ & t=0,\ v=20\ \Rightarrow \ c=20 \\ & \\ & v=20-\frac{k{{t}^{...)
- 15:31, 11 October 2010 (hist) (diff) Answer 19.5b (New page: 300 m)
- 15:29, 11 October 2010 (hist) (diff) Answer 19.5a (New page: <math>k=\frac{3}{64}</math>)
- 13:52, 11 October 2010 (hist) (diff) Answer 19.4c (New page: 693 m) (top)
- 13:52, 11 October 2010 (hist) (diff) Solution 19.4c (New page: Using the result from part b) for the velocity, <math>\begin{align} & s(20)=\int_{0}^{20}{v}dt \\ & \\ & =\int_{0}^{20}{\left( \frac{{{t}^{2}}}{5}+8 \right)dt} \\ & \\ & =\left[ \fr...) (top)
- 13:48, 11 October 2010 (hist) (diff) Solution 19.4b (New page: Using the result from part a) for the acceleration, <math>\begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( \frac{2t}{5} \right)}dt \\ & \\ & =\frac{{{t}^{2}}}{5}+c \\ & \\ & t=...) (top)
- 13:46, 11 October 2010 (hist) (diff) Answer 19.4b (New page: <math>\frac{{{t}^{2}}}{5}+8</math>) (top)
- 13:44, 11 October 2010 (hist) (diff) Solution 19.4a (New page: As the intercept is 0 and the gradient is <math>\frac{8000}{20}=400</math> , using the equation for a straight line, the force at a time <math>t</math> is given by: <math>F=400t+0=400t<...) (top)
- 13:37, 11 October 2010 (hist) (diff) 19. Exercises
- 13:35, 11 October 2010 (hist) (diff) Image:E19.4new.GIF (top)
- 09:36, 11 October 2010 (hist) (diff) Solution 19.3c (top)
- 09:36, 11 October 2010 (hist) (diff) Solution 19.2b (top)
- 09:35, 11 October 2010 (hist) (diff) Solution 19.1c (top)
- 09:33, 11 October 2010 (hist) (diff) Solution 19.3c (New page: <math>\begin{align} & s(10)=\int_{0}^{10}{v}dt \\ & \\ & =\int_{0}^{10}{\left( 8t-\frac{2{{t}^{2}}}{5} \right)dt} \\ & \\ & =\left[ 4{{t}^{2}}-\frac{2{{t}^{3}}}{15} \right]_{0}^{10} ...)
- 09:31, 11 October 2010 (hist) (diff) Answer 19.3c (New page: 267 m) (top)
- 09:31, 11 October 2010 (hist) (diff) Solution 19.3b (New page: <math>\begin{align} & v=\int{a}dt \\ & \\ & =\int{\left( 8-\frac{4t}{5} \right)}dt \\ & \\ & =8t-\frac{2{{t}^{2}}}{5}+c \\ & \\ & t=0,\ v=0\ \Rightarrow \ c=0 \\ & \\ & v=8t-\f...) (top)
- 09:26, 11 October 2010 (hist) (diff) m Answer 19.3b (New page: <math>40\text{ m}{{\text{s}}^{\text{-1}}}</math>) (top)
- 09:23, 11 October 2010 (hist) (diff) 19. Exercises
- 09:22, 11 October 2010 (hist) (diff) Solution 19.3a (New page: The graph below shows how the acceleration varies. Image:19.3a.gif As the intercept is 8 and the gradient is <math>-\frac{8}{10}=-\frac{4}{5}</math> , the acceleration is given by...) (top)
- 09:21, 11 October 2010 (hist) (diff) Image:19.3a.gif (top)
- 09:17, 11 October 2010 (hist) (diff) Solution 19.2b (New page: <math>\begin{align} & s(50)=\int_{0}^{50}{v}dt \\ & \\ & =\int_{0}^{50}{\left( \frac{{{t}^{2}}}{50}+2 \right)dt} \\ & \\ & =\left[ \frac{{{t}^{3}}}{150}+2t \right]_{0}^{50} \\ & \\...)
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