19. Variable acceleration II

Förberedande Mekanik

Nuvarande version

Key Points

Don't forget to evaluate constants of integration.

Example 19.1

The acceleration, \displaystyle a , of a particle, at time \displaystyle 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} \text{ m}{{\text{s}}^{\text{-1}}}

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}\text{ m}

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 \displaystyle 0\textrm{.}5\mathbf{j} \text{ m}{{\text{s}}^{-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\textrm{.}5\mathbf{j}. These values can be substituted to give \displaystyle {{c}_{1}}=0 and \displaystyle {{c}_{2}}=0\textrm{.}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\textrm{.}5 \right)\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}}

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\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} \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\textrm{.}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\textrm{.}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}) \displaystyle \text{m}{{\text{s}}^{-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} \text{ m}{{\text{s}}^{\text{-2}}} \\ \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} \text{ m}{{\text{s}}^{\text{-1}}}

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} \text{ m}