19. Variable acceleration II

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 +
 +
Key Points
 +
 +
 +
<math>\begin{align}
 +
& v=\int{a}dt \\
 +
& s=\int{v}dt \\
 +
\end{align}</math>
 +
 +
<math>\begin{align}
 +
& \mathbf{v}=\int{\mathbf{a}}dt \\
 +
& \mathbf{r}=\int{\mathbf{v}}dt \\
 +
\end{align}</math>
 +
 +
Don't forget to evaluate constants of integration.
 +
 +
 +
 +
Example 19.1
 +
 +
The acceleration,
 +
<math>a</math>
 +
<math>\text{m}{{\text{s}}^{-1}}</math>, of a particle, at time
 +
<math>t</math>
 +
t seconds is given by:
 +
 +
<math>a=4-\frac{t}{20}</math>
 +
<math>\text{m}{{\text{s}}^{-2}}</math>.
 +
 +
This model is valid for
 +
<math>0\le t\le 80</math>.
 +
Given that the particle starts at rest, find the distance travelled by the particle when
 +
<math>t=\text{ 8}0</math>
 +
.
 +
 +
Solution
 +
 +
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>
 +
 +
 +
To find the value of the constant
 +
<math>{{c}_{1}}</math>
 +
, note that the particle is initially at rest, so that
 +
<math>v=0</math>
 +
when
 +
<math>t=0</math>.
 +
Substituting these values shows that
 +
<math>{{c}_{1}}=0</math>.
 +
Hence the velocity is:
 +
 +
 +
<math>v=4t-\frac{{{t}^{2}}}{40}</math>
 +
 +
 +
 +
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>
 +
 +
 +
To find the constant
 +
<math>{{c}_{2}}</math>
 +
, assume that the particle starts at the origin, so that
 +
<math>s=0</math>
 +
when
 +
<math>t=0</math>.
 +
Hence
 +
<math>{{c}_{2}}=0</math>
 +
and the displacement at time
 +
<math>t</math>
 +
is given by:
 +
 +
 +
<math>s=2{{t}^{2}}-\frac{{{t}^{3}}}{120}</math>
 +
 +
 +
To find the distance travelled substitute
 +
<math>t=80</math>.
 +
 +
<math>s=2\times {{80}^{2}}-\frac{{{80}^{3}}}{120}=8530\text{ m (to 3sf)}</math>
 +
 +
 +
Example 19.2
 +
 +
A boat has an initial velocity of 0.5j ms-1 and experiences an acceleration of
 +
<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>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.
 +
 +
a) Find the velocity of the boat at time
 +
<math>t</math>
 +
.
 +
 +
b) Find the distance of the boat from the origin when
 +
<math>t=\text{ 12}0</math>
 +
.
 +
 +
Solution
 +
(a) Integrate the acceleration to obtain the velocity:
 +
 +
 +
<math>\begin{align}
 +
& \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}
 +
\end{align}</math>
 +
 +
 +
When
 +
<math>t=0</math>,
 +
 +
<math>\mathbf{v}=0.5\mathbf{j}</math>.
 +
These values can be substituted to give
 +
<math>{{c}_{1}}=0</math>
 +
and
 +
<math>{{c}_{2}}=0.5</math>,
 +
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>
 +
 +
 +
(b) Integrating the velocity gives the position vector:
 +
 +
 +
<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} \\
 +
& =\left( \frac{3{{t}^{2}}}{20}+{{c}_{3}} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0.5t+{{c}_{4}} \right)\mathbf{j}
 +
\end{align}</math>
 +
 +
 +
The boat is initially at the origin, so when
 +
<math>t=0</math>,
 +
the boat has position vector
 +
<math>0\mathbf{i}+0\mathbf{j}</math>.
 +
Substituting these values gives
 +
<math>{{c}_{3}}=0</math>
 +
and
 +
<math>{{c}_{4}}=0</math>.
 +
Hence the position vector is:
 +
 +
 +
<math>\mathbf{r}=\left( \frac{3{{t}^{2}}}{20} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0.5t \right)\mathbf{j}</math>
 +
 +
 +
Substituting
 +
<math>t=\text{12}0</math>
 +
, gives the required position vector.
 +
 +
 +
<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>
 +
 +
The distance from the origin can now be calculated, by finding the magnitude of the position vector.
 +
 +
 +
<math>s=\sqrt{{{2160}^{2}}+{{5820}^{2}}}=6210\text{ m (to 3sf)}</math>
 +
 +
 +
Example 19.3
 +
 +
At time
 +
<math>t</math>
 +
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>) ms-1.
 +
 +
a) Find the acceleration at time
 +
<math>t</math>
 +
.
 +
 +
b) Find the velocity of the particle at time
 +
<math>t</math>
 +
.
 +
 +
c) Find the position vector of the particle at time
 +
<math>t</math>
 +
.
 +
 +
Solution
 +
(a) Using Newton’s second Law,
 +
<math>\mathbf{F}=m\mathbf{a}</math>,
 +
gives:
 +
 +
 +
<math>\begin{align}
 +
& 300t\mathbf{i}-400t\mathbf{j}=250\mathbf{a} \\
 +
& \mathbf{a}=\frac{6}{5}t\mathbf{i}-\frac{8}{5}t\mathbf{j} \\
 +
\end{align}</math>
 +
 +
 +
(b) Integrating the acceleration gives the velocity:
 +
 +
 +
<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}}} \\
 +
& =\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>
 +
 +
 +
As the initial velocity is
 +
<math>2\mathbf{i}-3\mathbf{j}</math>,
 +
this can be used with
 +
<math>t=0</math>,
 +
to show that
 +
<math>{{c}_{1}}=2</math>
 +
and
 +
<math>{{c}_{2}}=-3</math>.
 +
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>
 +
 +
 +
Integrating the velocity gives the position vector:
 +
 +
 +
<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} \\
 +
& =\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>
 +
 +
 +
Note that the particle is initially at the origin. So using
 +
<math>\mathbf{r}=0\mathbf{i}+0\mathbf{j}</math>
 +
when
 +
<math>t=0</math>,
 +
gives
 +
<math>{{c}_{3}}=0</math>
 +
and
 +
<math>{{c}_{4}}=0</math>.
 +
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>

Revision as of 11:19, 19 November 2009

       Theory          Exercises      

Key Points


\displaystyle \begin{align} & v=\int{a}dt \\ & s=\int{v}dt \\ \end{align}

\displaystyle \begin{align} & \mathbf{v}=\int{\mathbf{a}}dt \\ & \mathbf{r}=\int{\mathbf{v}}dt \\ \end{align}

Don't forget to evaluate constants of integration.


Example 19.1

The acceleration, \displaystyle a \displaystyle \text{m}{{\text{s}}^{-1}}, of a particle, at time \displaystyle t t seconds is given by:

\displaystyle a=4-\frac{t}{20} \displaystyle \text{m}{{\text{s}}^{-2}}.

This model is valid for \displaystyle 0\le t\le 80. Given that the particle starts at rest, find the distance travelled by the particle when \displaystyle t=\text{ 8}0 .

Solution

First integrate the acceleration to obtain the velocity.


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


To find the value of the constant \displaystyle {{c}_{1}} , note that the particle is initially at rest, so that \displaystyle v=0 when \displaystyle t=0. Substituting these values shows that \displaystyle {{c}_{1}}=0. Hence the velocity is:


\displaystyle v=4t-\frac{{{t}^{2}}}{40}


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


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


To find the constant \displaystyle {{c}_{2}} , assume that the particle starts at the origin, so that \displaystyle s=0 when \displaystyle t=0. Hence \displaystyle {{c}_{2}}=0 and the displacement at time \displaystyle t 

 is given by:


\displaystyle s=2{{t}^{2}}-\frac{{{t}^{3}}}{120}


To find the distance travelled substitute \displaystyle t=80.

\displaystyle s=2\times {{80}^{2}}-\frac{{{80}^{3}}}{120}=8530\text{ m (to 3sf)}


Example 19.2

A boat has an initial velocity of 0.5j ms-1 and experiences an acceleration of \displaystyle \left( \frac{3}{10}\mathbf{i}+\left( \frac{t}{50} \right)\mathbf{j} \right) \displaystyle \text{m}{{\text{s}}^{-2}}, at time \displaystyle t 

seconds. Assume that the boat is initially at the origin. The unit vector \displaystyle \mathbf{i} and \displaystyle \mathbf{j} are directed east and north respectively.

a) Find the velocity of the boat at time \displaystyle t .

b) Find the distance of the boat from the origin when \displaystyle t=\text{ 12}0 .

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


\displaystyle \begin{align} & \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} \end{align}


When \displaystyle t=0,

\displaystyle \mathbf{v}=0.5\mathbf{j}. These values can be substituted to give \displaystyle {{c}_{1}}=0 and \displaystyle {{c}_{2}}=0.5, 

so that the velocity is given by:


\displaystyle \mathbf{v}=\left( \frac{3}{10}t \right)\mathbf{i}+\left( \frac{{{t}^{2}}}{100}+0.5 \right)\mathbf{j}


(b) Integrating the velocity gives the position vector:


\displaystyle \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} \\ & =\left( \frac{3{{t}^{2}}}{20}+{{c}_{3}} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0.5t+{{c}_{4}} \right)\mathbf{j} \end{align}


The boat is initially at the origin, so when \displaystyle t=0, 

the boat has position vector

\displaystyle 0\mathbf{i}+0\mathbf{j}. 

Substituting these values gives

\displaystyle {{c}_{3}}=0 and \displaystyle {{c}_{4}}=0. 

Hence the  position vector is:


\displaystyle \mathbf{r}=\left( \frac{3{{t}^{2}}}{20} \right)\mathbf{i}+\left( \frac{{{t}^{3}}}{300}+0.5t \right)\mathbf{j}


Substituting \displaystyle t=\text{12}0 , gives the required position vector.


\displaystyle \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}

The distance from the origin can now be calculated, by finding the magnitude of the position vector.


\displaystyle s=\sqrt{{{2160}^{2}}+{{5820}^{2}}}=6210\text{ m (to 3sf)}


Example 19.3

At time \displaystyle t 

seconds the resultant force on a particle, of mass 250 kg is

\displaystyle \left( 300t\mathbf{i}-400t\mathbf{j} \right) N. Initially the particle is at the origin and is moving with velocity (2\displaystyle \mathbf{i} - 3\displaystyle \mathbf{j}) ms-1.

a) Find the acceleration at time \displaystyle t .

b) Find the velocity of the particle at time \displaystyle t .

c) Find the position vector of the particle at time \displaystyle t .

Solution (a) Using Newton’s second Law, \displaystyle \mathbf{F}=m\mathbf{a}, gives:


\displaystyle \begin{align} & 300t\mathbf{i}-400t\mathbf{j}=250\mathbf{a} \\ & \mathbf{a}=\frac{6}{5}t\mathbf{i}-\frac{8}{5}t\mathbf{j} \\ \end{align}


(b) Integrating the acceleration gives the velocity:


\displaystyle \begin{align} & \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} \end{align}


As the initial velocity is \displaystyle 2\mathbf{i}-3\mathbf{j}, 

this can be used with

\displaystyle t=0, 

to show that

\displaystyle {{c}_{1}}=2 and \displaystyle {{c}_{2}}=-3. 

Hence the velocity is given by:


\displaystyle \mathbf{v}=\left( \frac{3}{5}{{t}^{2}}+2 \right)\mathbf{i}+\left( -\frac{4{{t}^{2}}}{5}-3 \right)\mathbf{j}


Integrating the velocity gives the position vector:


\displaystyle \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} \\ & =\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}


Note that the particle is initially at the origin. So using \displaystyle \mathbf{r}=0\mathbf{i}+0\mathbf{j} when \displaystyle t=0, 

gives

\displaystyle {{c}_{3}}=0 and \displaystyle {{c}_{4}}=0. 

Hence the position vector is given by:


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

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