19. Variable acceleration II

From Mechanics

Revision as of 11:38, 19 November 2009

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}

c) 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}