19. Variable acceleration II - Versionshistorik

Ian den 20 februari 2010 kl. 13.15

2010-02-20T13:15:41Z

← Äldre version		Versionen från 20 februari 2010 kl. 13.15
Rad 104:		Rad 104:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{v}=\int{\frac{3}{10}dt\mathbf{i}+\int{\left( \frac{t}{50} \right)dt\mathbf{j}}} \\	+	& \mathbf{v}=\int{\frac{3}{10}dt\ \mathbf{i}+\int{\left( \frac{t}{50} \right)dt\ \mathbf{j}}} \\
	& =\left( \frac{3}{10}t+{{c}_{1}} \right)\mathbf{i}+\left( \frac{{{t}^{\ 2}}}{100}+{{c}_{2}} \right)\mathbf{j}		& =\left( \frac{3}{10}t+{{c}_{1}} \right)\mathbf{i}+\left( \frac{{{t}^{\ 2}}}{100}+{{c}_{2}} \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>
Rad 123:		Rad 123:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{r}=\int{\left( \frac{3}{10}t \right)dt}\mathbf{i}+\int{\left( \frac{{{t}^{\ 2}}}{100}+0\textrm{.}5 \right)}dt\mathbf{j} \\	+	& \mathbf{r}=\int{\left( \frac{3}{10}t \right)dt}\ \mathbf{i}+\int{\left( \frac{{{t}^{\ 2}}}{100}+0\textrm{.}5 \right)}dt\ \mathbf{j} \\
	& =\left( \frac{3{{t}^{\ 2}}}{20}+{{c}_{3}} \right)\mathbf{i}+\left( \frac{{{t}^{\ 3}}}{300}+0\textrm{.}5t+{{c}_{4}} \right)\mathbf{j}		& =\left( \frac{3{{t}^{\ 2}}}{20}+{{c}_{3}} \right)\mathbf{i}+\left( \frac{{{t}^{\ 3}}}{300}+0\textrm{.}5t+{{c}_{4}} \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>
Rad 155:		Rad 155:
	<math>t</math>		<math>t</math>
	seconds the resultant force on a particle, of mass 250 kg is		seconds the resultant force on a particle, of mass 250 kg is
-	<math>\left( 300t\mathbf{i}-400t\mathbf{j} \right)</math>N. Initially the particle is at the origin and is moving with velocity (2<math>\mathbf{i}</math> - 3<math>\mathbf{j}</math>) <math>\text{m}{{\text{s}}^{-1}}</math>.	+	<math>\left( 300t\ \mathbf{i}-400t\ \mathbf{j} \right)</math>N. Initially the particle is at the origin and is moving with velocity (2<math>\mathbf{i}</math> - 3<math>\mathbf{j}</math>) <math>\text{m}{{\text{s}}^{-1}}</math>.

	a) Find the acceleration at time		a) Find the acceleration at time
Rad 173:		Rad 173:

	<math>\begin{align}		<math>\begin{align}
-	& 300t\mathbf{i}-400t\mathbf{j}=250\mathbf{a} \\	+	& 300t\ \mathbf{i}-400t\ \mathbf{j}=250\mathbf{a} \\
-	& \mathbf{a}=\frac{6}{5}t\mathbf{i}-\frac{8}{5}t\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}} \\	+	& \mathbf{a}=\frac{6}{5}t\ \mathbf{i}-\frac{8}{5}t\ \mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}} \\
	\end{align}</math>		\end{align}</math>

Rad 180:		Rad 180:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{v}=\int{\left( \frac{6}{5}t \right)dt\mathbf{i}+\int{\left( -\frac{8}{5}t \right)dt\mathbf{j}}} \\	+	& \mathbf{v}=\int{\left( \frac{6}{5}t \right)dt\ \mathbf{i}+\int{\left( -\frac{8}{5}t \right)dt\ \mathbf{j}}} \\
	& =\left( \frac{3}{5}{{t}^{\ 2}}+{{c}_{1}} \right)\mathbf{i}+\left( -\frac{4{{t}^{\ 2}}}{5}+{{c}_{2}} \right)\mathbf{j}		& =\left( \frac{3}{5}{{t}^{\ 2}}+{{c}_{1}} \right)\mathbf{i}+\left( -\frac{4{{t}^{\ 2}}}{5}+{{c}_{2}} \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>
Rad 199:		Rad 199:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{r}=\int{\left( \frac{3}{5}{{t}^{\ 2}}+2 \right)}dt\mathbf{i}+\int{\left( -\frac{4{{t}^{\ 2}}}{5}-3 \right)}dt\mathbf{j} \\	+	& \mathbf{r}=\int{\left( \frac{3}{5}{{t}^{\ 2}}+2 \right)}dt\ \mathbf{i}+\int{\left( -\frac{4{{t}^{\ 2}}}{5}-3 \right)}dt\ \mathbf{j} \\
	& =\left( \frac{{{t}^{\ 3}}}{5}+2t+{{c}_{3}} \right)\mathbf{i}+\left( -\frac{4{{t}^{\ 3}}}{15}-3t+{{c}_{4}} \right)\mathbf{j}		& =\left( \frac{{{t}^{\ 3}}}{5}+2t+{{c}_{3}} \right)\mathbf{i}+\left( -\frac{4{{t}^{\ 3}}}{15}-3t+{{c}_{4}} \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>

Ian den 20 februari 2010 kl. 13.02

2010-02-20T13:02:55Z

← Äldre version		Versionen från 20 februari 2010 kl. 13.02
Rad 47:		Rad 47:
	First integrate the acceleration to obtain the velocity.		First integrate the acceleration to obtain the velocity.

-	<math>v=\int{\left( 4-\frac{t}{20} \right)}dt=4t-\frac{{{t}^{2}}}{40}+{{c}_{1}}</math>	+	<math>v=\int{\left( 4-\frac{t}{20} \right)}dt=4t-\frac{{{t}^{\ 2}}}{40}+{{c}_{1}}</math>

	To find the value of the constant		To find the value of the constant
Rad 59:		Rad 59:
	Hence the velocity is:		Hence the velocity is:

-	<math>v=4t-\frac{{{t}^{2}}}{40} \text{ m}{{\text{s}}^{\text{-1}}}</math>	+	<math>v=4t-\frac{{{t}^{\ 2}}}{40} \text{ m}{{\text{s}}^{\text{-1}}}</math>

	The displacement of the particle can be found by integrating the velocity:		The displacement of the particle can be found by integrating the velocity:

-	<math>s=\int{\left( 4t-\frac{{{t}^{2}}}{40} \right)}dt=2{{t}^{2}}-\frac{{{t}^{3}}}{120}+{{c}_{2}}</math>	+	<math>s=\int{\left( 4t-\frac{{{t}^{\ 2}}}{40} \right)}dt=2{{t}^{\ 2}}-\frac{{{t}^{\ 3}}}{120}+{{c}_{2}}</math>

	To find the constant		To find the constant
Rad 77:		Rad 77:
	is given by:		is given by:

-	<math>s=2{{t}^{2}}-\frac{{{t}^{3}}}{120}\text{ m}</math>	+	<math>s=2{{t}^{\ 2}}-\frac{{{t}^{\ 3}}}{120}\text{ m}</math>

	To find the distance travelled substitute		To find the distance travelled substitute
Rad 105:		Rad 105:
	<math>\begin{align}		<math>\begin{align}
	& \mathbf{v}=\int{\frac{3}{10}dt\mathbf{i}+\int{\left( \frac{t}{50} \right)dt\mathbf{j}}} \\		& \mathbf{v}=\int{\frac{3}{10}dt\mathbf{i}+\int{\left( \frac{t}{50} \right)dt\mathbf{j}}} \\
-	& =\left( \frac{3}{10}t+{{c}_{1}} \right)\mathbf{i}+\left( \frac{{{t}^{2}}}{100}+{{c}_{2}} \right)\mathbf{j}	+	& =\left( \frac{3}{10}t+{{c}_{1}} \right)\mathbf{i}+\left( \frac{{{t}^{\ 2}}}{100}+{{c}_{2}} \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>

Rad 118:		Rad 118:
	so that the velocity is given by:		so that the velocity is given by:

-	<math>\mathbf{v}=\left( \frac{3}{10}t \right)\mathbf{i}+\left( \frac{{{t}^{2}}}{100}+0\textrm{.}5 \right)\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}</math>	+	<math>\mathbf{v}=\left( \frac{3}{10}t \right)\mathbf{i}+\left( \frac{{{t}^{\ 2}}}{100}+0\textrm{.}5 \right)\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}</math>

	b) Integrating the velocity gives the position vector:		b) Integrating the velocity gives the position vector:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{r}=\int{\left( \frac{3}{10}t \right)dt}\mathbf{i}+\int{\left( \frac{{{t}^{2}}}{100}+0\textrm{.}5 \right)}dt\mathbf{j} \\	+	& \mathbf{r}=\int{\left( \frac{3}{10}t \right)dt}\mathbf{i}+\int{\left( \frac{{{t}^{\ 2}}}{100}+0\textrm{.}5 \right)}dt\mathbf{j} \\
-	& =\left( \frac{3{{t}^{2}}}{20}+{{c}_{3}} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0\textrm{.}5t+{{c}_{4}} \right)\mathbf{j}	+	& =\left( \frac{3{{t}^{\ 2}}}{20}+{{c}_{3}} \right)\mathbf{i}+\left( \frac{{{t}^{\ 3}}}{300}+0\textrm{.}5t+{{c}_{4}} \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>

Rad 137:		Rad 137:
	Hence the position vector is:		Hence the position vector is:

-	<math>\mathbf{r}=\left( \frac{3{{t}^{2}}}{20} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0\textrm{.}5t \right)\mathbf{j}</math>	+	<math>\mathbf{r}=\left( \frac{3{{t}^{\ 2}}}{20} \right)\mathbf{i}+\left( \frac{{{t}^{\ 3}}}{300}+0\textrm{.}5t \right)\mathbf{j}</math>

	Substituting		Substituting
Rad 143:		Rad 143:
	, gives the required position vector.		, gives the required position vector.

-	<math>\mathbf{r}=\left( \frac{3\times {{120}^{2}}}{20} \right)\mathbf{i}+\left( \frac{{{120}^{3}}}{300}+0\textrm{.}5\times 120 \right)\mathbf{j}=2160\mathbf{i}+5820\mathbf{j}</math>	+	<math>\mathbf{r}=\left( \frac{3\times {{120}^{\ 2}}}{20} \right)\mathbf{i}+\left( \frac{{{120}^{3}}}{300}+0\textrm{.}5\times 120 \right)\mathbf{j}=2160\mathbf{i}+5820\mathbf{j}</math>

	The distance from the origin can now be calculated, by finding the magnitude of the position vector.		The distance from the origin can now be calculated, by finding the magnitude of the position vector.
Rad 181:		Rad 181:
	<math>\begin{align}		<math>\begin{align}
	& \mathbf{v}=\int{\left( \frac{6}{5}t \right)dt\mathbf{i}+\int{\left( -\frac{8}{5}t \right)dt\mathbf{j}}} \\		& \mathbf{v}=\int{\left( \frac{6}{5}t \right)dt\mathbf{i}+\int{\left( -\frac{8}{5}t \right)dt\mathbf{j}}} \\
-	& =\left( \frac{3}{5}{{t}^{2}}+{{c}_{1}} \right)\mathbf{i}+\left( -\frac{4{{t}^{2}}}{5}+{{c}_{2}} \right)\mathbf{j}	+	& =\left( \frac{3}{5}{{t}^{\ 2}}+{{c}_{1}} \right)\mathbf{i}+\left( -\frac{4{{t}^{\ 2}}}{5}+{{c}_{2}} \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>

Rad 194:		Rad 194:
	Hence the velocity is given by		Hence the velocity is given by

-	<math>\mathbf{v}=\left( \frac{3}{5}{{t}^{2}}+2 \right)\mathbf{i}+\left( -\frac{4{{t}^{2}}}{5}-3 \right)\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}</math>	+	<math>\mathbf{v}=\left( \frac{3}{5}{{t}^{\ 2}}+2 \right)\mathbf{i}+\left( -\frac{4{{t}^{\ 2}}}{5}-3 \right)\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}</math>

	c) Integrating the velocity gives the position vector:		c) Integrating the velocity gives the position vector:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{r}=\int{\left( \frac{3}{5}{{t}^{2}}+2 \right)}dt\mathbf{i}+\int{\left( -\frac{4{{t}^{2}}}{5}-3 \right)}dt\mathbf{j} \\	+	& \mathbf{r}=\int{\left( \frac{3}{5}{{t}^{\ 2}}+2 \right)}dt\mathbf{i}+\int{\left( -\frac{4{{t}^{\ 2}}}{5}-3 \right)}dt\mathbf{j} \\
-	& =\left( \frac{{{t}^{3}}}{5}+2t+{{c}_{3}} \right)\mathbf{i}+\left( -\frac{4{{t}^{3}}}{15}-3t+{{c}_{4}} \right)\mathbf{j}	+	& =\left( \frac{{{t}^{\ 3}}}{5}+2t+{{c}_{3}} \right)\mathbf{i}+\left( -\frac{4{{t}^{\ 3}}}{15}-3t+{{c}_{4}} \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>

Rad 213:		Rad 213:
	Hence the position vector is given by:		Hence the position vector is given by:

-	<math>\mathbf{r}=\left( \frac{{{t}^{3}}}{5}+2t \right)\mathbf{i}+\left( -\frac{4{{t}^{3}}}{15}-3t \right)\mathbf{j} \text{ m}</math>	+	<math>\mathbf{r}=\left( \frac{{{t}^{\ 3}}}{5}+2t \right)\mathbf{i}+\left( -\frac{4{{t}^{\ 3}}}{15}-3t \right)\mathbf{j} \text{ m}</math>

Ian den 20 februari 2010 kl. 12.55

2010-02-20T12:55:19Z

← Äldre version		Versionen från 20 februari 2010 kl. 12.55
Rad 118:		Rad 118:
	so that the velocity is given by:		so that the velocity is given by:

-	<math>\mathbf{v}=\left( \frac{3}{10}t \right)\mathbf{i}+\left( \frac{{{t}^{2}}}{100}+\textrm{.}5 \right)\mathbf{j}</math>	+	<math>\mathbf{v}=\left( \frac{3}{10}t \right)\mathbf{i}+\left( \frac{{{t}^{2}}}{100}+0\textrm{.}5 \right)\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}</math>

	b) Integrating the velocity gives the position vector:		b) Integrating the velocity gives the position vector:
Rad 155:		Rad 155:
	<math>t</math>		<math>t</math>
	seconds the resultant force on a particle, of mass 250 kg is		seconds the resultant force on a particle, of mass 250 kg is
-	<math>\left( 300t\mathbf{i}-400t\mathbf{j} \right)</math>	+	<math>\left( 300t\mathbf{i}-400t\mathbf{j} \right)</math>N. Initially the particle is at the origin and is moving with velocity (2<math>\mathbf{i}</math> - 3<math>\mathbf{j}</math>) <math>\text{m}{{\text{s}}^{-1}}</math>.
-	N. Initially the particle is at the origin and is moving with velocity (2<math>\mathbf{i}</math> - 3<math>\mathbf{j}</math>) <math>\text{m}{{\text{s}}^{-1}}</math>.	+

	a) Find the acceleration at time		a) Find the acceleration at time
Rad 175:		Rad 174:
	<math>\begin{align}		<math>\begin{align}
	& 300t\mathbf{i}-400t\mathbf{j}=250\mathbf{a} \\		& 300t\mathbf{i}-400t\mathbf{j}=250\mathbf{a} \\
-	& \mathbf{a}=\frac{6}{5}t\mathbf{i}-\frac{8}{5}t\mathbf{j} \\	+	& \mathbf{a}=\frac{6}{5}t\mathbf{i}-\frac{8}{5}t\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}} \\
	\end{align}</math>		\end{align}</math>

Rad 195:		Rad 194:
	Hence the velocity is given by		Hence the velocity is given by

-	<math>\mathbf{v}=\left( \frac{3}{5}{{t}^{2}}+2 \right)\mathbf{i}+\left( -\frac{4{{t}^{2}}}{5}-3 \right)\mathbf{j}</math>	+	<math>\mathbf{v}=\left( \frac{3}{5}{{t}^{2}}+2 \right)\mathbf{i}+\left( -\frac{4{{t}^{2}}}{5}-3 \right)\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}</math>

	c) Integrating the velocity gives the position vector:		c) Integrating the velocity gives the position vector:
Rad 214:		Rad 213:
	Hence the position vector is given by:		Hence the position vector is given by:

-	<math>\mathbf{r}=\left( \frac{{{t}^{3}}}{5}+2t \right)\mathbf{i}+\left( -\frac{4{{t}^{3}}}{15}-3t \right)\mathbf{j}</math>	+	<math>\mathbf{r}=\left( \frac{{{t}^{3}}}{5}+2t \right)\mathbf{i}+\left( -\frac{4{{t}^{3}}}{15}-3t \right)\mathbf{j} \text{ m}</math>

Ian den 20 februari 2010 kl. 12.31

2010-02-20T12:31:34Z

(Skillnad mellan versioner)

Ian den 19 november 2009 kl. 12.20

2009-11-19T12:20:37Z

← Äldre version		Versionen från 19 november 2009 kl. 12.20
Rad 97:		Rad 97:
	'''[[Example 19.2]]'''		'''[[Example 19.2]]'''

-	A boat has an initial velocity of 0.5j ms-1 and experiences an acceleration of	+	A boat has an initial velocity of 0\textrm{.}5j <math>\text{m}{{\text{s}}^{-1}}</math> and experiences an acceleration of
	<math>\left( \frac{3}{10}\mathbf{i}+\left( \frac{t}{50} \right)\mathbf{j} \right)</math>		<math>\left( \frac{3}{10}\mathbf{i}+\left( \frac{t}{50} \right)\mathbf{j} \right)</math>
	<math>\text{m}{{\text{s}}^{-2}}</math>, at time		<math>\text{m}{{\text{s}}^{-2}}</math>, at time
Rad 125:		Rad 125:
	<math>t=0</math>,		<math>t=0</math>,

-	<math>\mathbf{v}=0.5\mathbf{j}</math>.	+	<math>\mathbf{v}=0\textrm{.}5\mathbf{j}</math>.
	These values can be substituted to give		These values can be substituted to give
	<math>{{c}_{1}}=0</math>		<math>{{c}_{1}}=0</math>
	and		and
-	<math>{{c}_{2}}=0.5</math>,	+	<math>{{c}_{2}}=0\textrm{.}5</math>,
	so that the velocity is given by:		so that the velocity is given by:


-	<math>\mathbf{v}=\left( \frac{3}{10}t \right)\mathbf{i}+\left( \frac{{{t}^{2}}}{100}+0.5 \right)\mathbf{j}</math>	+	<math>\mathbf{v}=\left( \frac{3}{10}t \right)\mathbf{i}+\left( \frac{{{t}^{2}}}{100}+\textrm{.}5 \right)\mathbf{j}</math>


Rad 140:		Rad 140:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{r}=\int{\left( \frac{3}{10}t \right)dt}\mathbf{i}+\int{\left( \frac{{{t}^{2}}}{100}+0.5 \right)}dt\mathbf{j} \\	+	& \mathbf{r}=\int{\left( \frac{3}{10}t \right)dt}\mathbf{i}+\int{\left( \frac{{{t}^{2}}}{100}+0\textrm{.}5 \right)}dt\mathbf{j} \\
-	& =\left( \frac{3{{t}^{2}}}{20}+{{c}_{3}} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0.5t+{{c}_{4}} \right)\mathbf{j}	+	& =\left( \frac{3{{t}^{2}}}{20}+{{c}_{3}} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0\textrm{.}5t+{{c}_{4}} \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>

Rad 156:		Rad 156:


-	<math>\mathbf{r}=\left( \frac{3{{t}^{2}}}{20} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0.5t \right)\mathbf{j}</math>	+	<math>\mathbf{r}=\left( \frac{3{{t}^{2}}}{20} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0\textrm{.}5t \right)\mathbf{j}</math>


Rad 164:		Rad 164:


-	<math>\mathbf{r}=\left( \frac{3\times {{120}^{2}}}{20} \right)\mathbf{i}+\left( \frac{{{120}^{3}}}{300}+0.5\times 120 \right)\mathbf{j}=2160\mathbf{i}+5820\mathbf{j}</math>	+	<math>\mathbf{r}=\left( \frac{3\times {{120}^{2}}}{20} \right)\mathbf{i}+\left( \frac{{{120}^{3}}}{300}+0\textrm{.}5\times 120 \right)\mathbf{j}=2160\mathbf{i}+5820\mathbf{j}</math>

	The distance from the origin can now be calculated, by finding the magnitude of the position vector.		The distance from the origin can now be calculated, by finding the magnitude of the position vector.
Rad 178:		Rad 178:
	seconds the resultant force on a particle, of mass 250 kg is		seconds the resultant force on a particle, of mass 250 kg is
	<math>\left( 300t\mathbf{i}-400t\mathbf{j} \right)</math>		<math>\left( 300t\mathbf{i}-400t\mathbf{j} \right)</math>
-	N. Initially the particle is at the origin and is moving with velocity (2<math>\mathbf{i}</math> - 3<math>\mathbf{j}</math>) ms-1.	+	N. Initially the particle is at the origin and is moving with velocity (2<math>\mathbf{i}</math> - 3<math>\mathbf{j}</math>) <math>\text{m}{{\text{s}}^{-1}}</math>.

	a) Find the acceleration at time		a) Find the acceleration at time

Ian den 19 november 2009 kl. 12.12

2009-11-19T12:12:38Z

← Äldre version		Versionen från 19 november 2009 kl. 12.12
Rad 12:		Rad 12:

	\|align="left"\| <math>\begin{align}		\|align="left"\| <math>\begin{align}
-	& v=\frac{ds}{dt} \\
-	& a=\frac{dv}{dt}=\frac{{{d}^{2}}s}{d{{t}^{2}}} \\
-	\end{align}</math>
-	\| valign="top"\|<math>\begin{align}
-	& \mathbf{r}=x\mathbf{i}+y\mathbf{j} \\
-	& \mathbf{v}=\frac{dx}{dt}\mathbf{i}+\frac{dy}{dt}\mathbf{j} \\
-	& \mathbf{a}=\frac{{{d}^{2}}x}{d{{t}^{2}}}\mathbf{i}+\frac{{{d}^{2}}y}{d{{t}^{2}}}\mathbf{j} \\
-	\end{align}</math>
-
-	\|}
-
-
-
-	<math>\begin{align}
	& v=\int{a}dt \\		& v=\int{a}dt \\
	& s=\int{v}dt \\		& s=\int{v}dt \\
	\end{align}</math>		\end{align}</math>
-		+	\| valign="top"\|<math>\begin{align}
-	<math>\begin{align}	+
	& \mathbf{v}=\int{\mathbf{a}}dt \\		& \mathbf{v}=\int{\mathbf{a}}dt \\
	& \mathbf{r}=\int{\mathbf{v}}dt \\		& \mathbf{r}=\int{\mathbf{v}}dt \\
	\end{align}</math>		\end{align}</math>

		+	\|}
		+
		+
	Don't forget to evaluate constants of integration.		Don't forget to evaluate constants of integration.

Ian den 19 november 2009 kl. 11.44

2009-11-19T11:44:13Z

← Äldre version		Versionen från 19 november 2009 kl. 11.44
Rad 8:		Rad 8:

	Key Points		Key Points
		+
		+	{\| width="100%" cellspacing="10px" align="center"
		+
		+	\|align="left"\| <math>\begin{align}
		+	& v=\frac{ds}{dt} \\
		+	& a=\frac{dv}{dt}=\frac{{{d}^{2}}s}{d{{t}^{2}}} \\
		+	\end{align}</math>
		+	\| valign="top"\|<math>\begin{align}
		+	& \mathbf{r}=x\mathbf{i}+y\mathbf{j} \\
		+	& \mathbf{v}=\frac{dx}{dt}\mathbf{i}+\frac{dy}{dt}\mathbf{j} \\
		+	& \mathbf{a}=\frac{{{d}^{2}}x}{d{{t}^{2}}}\mathbf{i}+\frac{{{d}^{2}}y}{d{{t}^{2}}}\mathbf{j} \\
		+	\end{align}</math>
		+
		+	\|}
		+


Rad 24:		Rad 39:


-	Example 19.1	+	'''[[Example 19.1]]'''

	The acceleration,		The acceleration,
Rad 41:		Rad 56:
	.		.

-	Solution	+	'''Solution'''

	First integrate the acceleration to obtain the velocity.		First integrate the acceleration to obtain the velocity.
Rad 92:		Rad 107:


-	Example 19.2	+	'''[[Example 19.2]]'''

	A boat has an initial velocity of 0.5j ms-1 and experiences an acceleration of		A boat has an initial velocity of 0.5j ms-1 and experiences an acceleration of
Rad 108:		Rad 123:
	.		.

-	Solution	+	'''Solution'''

	a) Integrate the acceleration to obtain the velocity:		a) Integrate the acceleration to obtain the velocity:
Rad 169:		Rad 184:


-	Example 19.3	+	'''[[Example 19.3]]'''

	At time		At time
Rad 188:		Rad 203:
	<math>t</math>.		<math>t</math>.

-	Solution	+	'''Solution'''

	a) Using Newton’s second Law,		a) Using Newton’s second Law,

Ian den 19 november 2009 kl. 11.38

2009-11-19T11:38:53Z

← Äldre version		Versionen från 19 november 2009 kl. 11.38
Rad 80:		Rad 80:
	and the displacement at time		and the displacement at time
	<math>t</math>		<math>t</math>
-	is given by:	+	is given by:


Rad 98:		Rad 98:
	<math>\text{m}{{\text{s}}^{-2}}</math>, at time		<math>\text{m}{{\text{s}}^{-2}}</math>, at time
	<math>t</math>		<math>t</math>
-	seconds. Assume that the boat is initially at the origin. The unit vector <math>\mathbf{i}</math> and <math>\mathbf{j}</math> are directed east and north respectively.	+	seconds. Assume that the boat is initially at the origin. The unit vector <math>\mathbf{i}</math> and <math>\mathbf{j}</math> are directed east and north respectively.

	a) Find the velocity of the boat at time		a) Find the velocity of the boat at time
Rad 109:		Rad 109:

	Solution		Solution
-	(a) Integrate the acceleration to obtain the velocity:	+
		+	a) Integrate the acceleration to obtain the velocity:


Rad 126:		Rad 127:
	and		and
	<math>{{c}_{2}}=0.5</math>,		<math>{{c}_{2}}=0.5</math>,
-	so that the velocity is given by:	+	so that the velocity is given by:


Rad 132:		Rad 133:


-	(b) Integrating the velocity gives the position vector:	+	b) Integrating the velocity gives the position vector:


Rad 143:		Rad 144:
	The boat is initially at the origin, so when		The boat is initially at the origin, so when
	<math>t=0</math>,		<math>t=0</math>,
-	the boat has position vector	+	the boat has position vector
	<math>0\mathbf{i}+0\mathbf{j}</math>.		<math>0\mathbf{i}+0\mathbf{j}</math>.
-	Substituting these values gives	+	Substituting these values gives
	<math>{{c}_{3}}=0</math>		<math>{{c}_{3}}=0</math>
	and		and
	<math>{{c}_{4}}=0</math>.		<math>{{c}_{4}}=0</math>.
-	Hence the position vector is:	+	Hence the position vector is:


Rad 172:		Rad 173:
	At time		At time
	<math>t</math>		<math>t</math>
-	seconds the resultant force on a particle, of mass 250 kg is	+	seconds the resultant force on a particle, of mass 250 kg is
	<math>\left( 300t\mathbf{i}-400t\mathbf{j} \right)</math>		<math>\left( 300t\mathbf{i}-400t\mathbf{j} \right)</math>
	N. Initially the particle is at the origin and is moving with velocity (2<math>\mathbf{i}</math> - 3<math>\mathbf{j}</math>) ms-1.		N. Initially the particle is at the origin and is moving with velocity (2<math>\mathbf{i}</math> - 3<math>\mathbf{j}</math>) ms-1.
Rad 185:		Rad 186:

	c) Find the position vector of the particle at time		c) Find the position vector of the particle at time
-	<math>t</math>	+	<math>t</math>.
-	.	+

	Solution		Solution
-	(a) Using Newton’s second Law,	+
		+	a) Using Newton’s second Law,
	<math>\mathbf{F}=m\mathbf{a}</math>,		<math>\mathbf{F}=m\mathbf{a}</math>,
	gives:		gives:
Rad 200:		Rad 201:


-	(b) Integrating the acceleration gives the velocity:	+	b) Integrating the acceleration gives the velocity:


Rad 211:		Rad 212:
	As the initial velocity is		As the initial velocity is
	<math>2\mathbf{i}-3\mathbf{j}</math>,		<math>2\mathbf{i}-3\mathbf{j}</math>,
-	this can be used with	+	this can be used with
	<math>t=0</math>,		<math>t=0</math>,
-	to show that	+	to show that
	<math>{{c}_{1}}=2</math>		<math>{{c}_{1}}=2</math>
	and		and
	<math>{{c}_{2}}=-3</math>.		<math>{{c}_{2}}=-3</math>.
-	Hence the velocity is given by:	+	Hence the velocity is given by:


Rad 223:		Rad 224:


-	Integrating the velocity gives the position vector:	+	c) Integrating the velocity gives the position vector:


Rad 236:		Rad 237:
	when		when
	<math>t=0</math>,		<math>t=0</math>,
-	gives	+	gives
	<math>{{c}_{3}}=0</math>		<math>{{c}_{3}}=0</math>
	and		and
	<math>{{c}_{4}}=0</math>.		<math>{{c}_{4}}=0</math>.
-	Hence the position vector is given by:	+	Hence the position vector is given by:


	<math>\mathbf{r}=\left( \frac{{{t}^{3}}}{5}+2t \right)\mathbf{i}+\left( -\frac{4{{t}^{3}}}{15}-3t \right)\mathbf{j}</math>		<math>\mathbf{r}=\left( \frac{{{t}^{3}}}{5}+2t \right)\mathbf{i}+\left( -\frac{4{{t}^{3}}}{15}-3t \right)\mathbf{j}</math>

Ian den 19 november 2009 kl. 11.19

2009-11-19T11:19:40Z

(Skillnad mellan versioner)

Ian: New page: NOTOC {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" | {{Selected tab|[[19. Variable acceleration I...

2009-11-16T18:33:38Z

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