Chapter 4, Problem 4P

Two forces, ${\stackrel{\to }{F}}_1=(-6\hat{i}-4\hat{j})\text{N}$ and ${\stackrel{\to }{F}}_2=(-3\hat{i}+7\hat{j})\text{N}$, act on a particle of mass 2.00 kg that is initially at rest at coordinates (−2.00 m, +4.00 m). (a) What are the components of the particle’s velocity at t = 10.0 s? (b) In what direction is the particle moving at t = 10.0 s? (c) What displacement does the particle undergo during the first 10.0 s? (d) What are the coordinates of the particle at t = 10.0 s?

To determine

The components of the particle’s velocity.

Answer to Problem 4P

The components of the particle’s velocity is $\underset{\_}{-45.0\text{m/s , 15}\text{.0 m/s}}$.

Explanation of Solution

Write the expression for velocity.

$\overrightarrow{v}=\overrightarrow{a}t$        (I)

Here, $\overrightarrow{v}$ is the velocity, $\overrightarrow{a}$ is the acceleration and $t$ is the time.

Write the expression for acceleration.

$\overrightarrow{a}=\frac{{\overrightarrow{F}}_{net}}{m}$        (II)

Here, ${\overrightarrow{F}}_{net}$ is the total force, $m$ is the mass.

Write the expression for net force.

${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2$        (III)

Here, ${\overrightarrow{F}}_1,{\overrightarrow{F}}_2$ are the applied forces.

Rewrite the expression for (I) by using (II) and (III).

$\overrightarrow{v}=\left(\frac{{\overrightarrow{F}}_1+{\overrightarrow{F}}_2}{m}\right)t$        (IV)

Conclusion:

Substitute, $\left(-6\widehat{i}-4\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_1$, $\left(-3\widehat{i}+7\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_2$, $2.00\text{kg}$ for $m$, $10.0\text{s}$ for $t$ in equation (IV) to find $\overrightarrow{v}$.

$\begin{array}{c}\overrightarrow{v}=\left(\frac{\left(\left(-6\widehat{i}-4\widehat{j}\right)\text{N}\right)+\left(\left(-3\widehat{i}+7\widehat{j}\right)\text{N}\right)}{\left(2.00\text{kg}\right)}\right)\left(10.0\text{s}\right)\\ =\left[\left(-4.50\widehat{i}+1.50\widehat{j}\right){\text{m/s}}^{\text{2}}\right]\left(10.0\text{s}\right)\\ =\left(-45.0\widehat{i}+15\widehat{j}\right)\text{m/s}\end{array}$

Thus, the components of the particle’s velocity is $\underset{\_}{-45.0\text{m/s , 15}\text{.0 m/s}}$.

To determine

The direction of the moving particle.

Answer to Problem 4P

The direction of the moving particle is $\underset{\_}{162°\text{ from the x-axis}}$.

Explanation of Solution

Write the expression for the direction of motion.

$\theta =180°+\left[{\tan }^{-1}\left(\frac{v_y}{v_x}\right)\right]$        (V)

Here, $\theta$ is the angle, $v_x,v_y$ are the x and y component of velocity.

Conclusion:

Substitute, $15.0\text{m/s}$ for $v_y$, $-45.0\text{m/s}$ for $v_x$ in equation (V) to find $\theta$.

$\begin{array}{c}\theta =180°+\left[{\tan }^{-1}\left(\frac{15.0\text{m/s}}{-45.0\text{m/s}}\right)\right]\\ =180°+\left(-18.4°\right)\\ \approx 162°\text{from the x-axis}\end{array}$

Thus, the direction of the moving particle is $\underset{\_}{162°\text{ from the x-axis}}$.

To determine

The displacement of the particle.

Answer to Problem 4P

The displacement of the particle is $\underset{\_}{\left(-225\widehat{i}+75.0\widehat{j}\right)\text{m}}$.

Explanation of Solution

Write the expression for displacement.

$\left(\Delta \overrightarrow{r}\right)=\left(\Delta \overrightarrow{x}\right)+\left(\Delta \overrightarrow{y}\right)$        (IV)

Here, $\left(\Delta \overrightarrow{r}\right)$ is the displacement, $\left(\Delta \overrightarrow{x}\right)$ is the x-displacement, $\left(\Delta \overrightarrow{y}\right)$ is the y-displacement.

Write the expression for x-displacement.

$\left(\Delta \overrightarrow{x}\right)={\overrightarrow{v}}_{xi}t+\frac{1}{2}\left(a_x\widehat{i}\right)t^2$        (V)

Here, ${\overrightarrow{v}}_{xi}$ is the initial velocity along x-axis, $t$ is the time, $a_x\widehat{i}$ is the x-component of acceleration.

Write the expression for y-displacement.

$\left(\Delta \overrightarrow{y}\right)={\overrightarrow{v}}_{yi}t+\frac{1}{2}\left(a_y\widehat{j}\right)t^2$        (VI)

Here, ${\overrightarrow{v}}_{yi}$ is the initial velocity along y-axis, $t$ is the time, $a_y\widehat{i}$ is the y-component of acceleration.

Write the expression for the acceleration.

$\overrightarrow{a}=a_x\widehat{i}+a_y\widehat{j}$        (VII)

Rewrite the expression for the displacement by using (II), (III), (V), (VI) and (VII).

$\begin{array}{c}\left(\Delta \overrightarrow{r}\right)=\left[{\overrightarrow{v}}_{xi}t+\frac{1}{2}\left(a_x\widehat{i}\right)t^2\right]+\left[{\overrightarrow{v}}_{yi}t+\frac{1}{2}\left(a_y\widehat{j}\right)t^2\right]\\ =\left({\overrightarrow{v}}_{xi}+{\overrightarrow{v}}_{yi}\right)t+\frac{1}{2}{t}^2\left(a_x\widehat{i}+a_y\widehat{j}\right)\\ =\left({\overrightarrow{v}}_{xi}+{\overrightarrow{v}}_{yi}\right)t+\frac{1}{2}{t}^2\overrightarrow{a}\\ =\left({\overrightarrow{v}}_{xi}+{\overrightarrow{v}}_{yi}\right)t+\frac{1}{2}{t}^2\left(\frac{{\overrightarrow{F}}_1+{\overrightarrow{F}}_2}{m}\right)\end{array}$        (VIII)

Conclusion:

Substitute, $\left(-6\widehat{i}-4\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_1$, $\left(-3\widehat{i}+7\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_2$, $2.00\text{kg}$ for $m$, $10.0\text{s}$ for $t$, $0$ for ${\overrightarrow{v}}_{xi}$, $0$ for ${\overrightarrow{v}}_{yi}$ in equation (VIII) to find $\left(\Delta \overrightarrow{r}\right)$.

$\begin{array}{c}\left(\Delta \overrightarrow{r}\right)=\left[\left(0+0\right)\left(10.0\text{s}\right)\right]+\left[\frac{1}{2}{\left(10.0\text{s}\right)}^2\left(\frac{\left[\left(-6\widehat{i}-4\widehat{j}\right)\text{N}\right]+\left[\left(-3\widehat{i}+7\widehat{j}\right)\text{N}\right]}{\left(2.00\text{kg}\right)}\right)\right]\\ =\left[\frac{1}{2}{\left(10.0\text{s}\right)}^2\left(\frac{\left[\left(-6\widehat{i}-4\widehat{j}\right)\text{N}\right]+\left[\left(-3\widehat{i}+7\widehat{j}\right)\text{N}\right]}{\left(2.00\text{kg}\right)}\right)\right]\\ =\left(-225\widehat{i}+75.0\widehat{j}\right)\text{m}\end{array}$

Thus, the displacement of the particle is $\underset{\_}{\left(-225\widehat{i}+75.0\widehat{j}\right)\text{m}}$.

To determine

The coordinates of the particle.

Answer to Problem 4P

The coordinates of the particle is $\underset{\_}{\left(-227\text{m , }79.0\text{m}\right)}$.

Explanation of Solution

Write the expression for position of the particle.

${\overrightarrow{r}}_f={\overrightarrow{r}}_i+\left(\Delta \overrightarrow{r}\right)$        (IX)

Here, ${\overrightarrow{r}}_f$ is the final position, ${\overrightarrow{r}}_i$ is the initial position.

Conclusion:

Substitute, $\left(-2.00\widehat{\text{i}}\text{+}4.00\widehat{\text{j}}\right)\text{m}$ for ${\overrightarrow{r}}_i$, $\left(-225\widehat{\text{i}}+75\widehat{\text{j}}\right)\text{m}$ for $\left(\Delta \overrightarrow{r}\right)$ in equation (IX) to find ${\overrightarrow{r}}_f$.

$\begin{array}{c}{\overrightarrow{r}}_f=\left(\left(-2.00\widehat{\text{i}}\text{+}4.00\widehat{\text{j}}\right)\text{m}\right)+\left(\left(-225\widehat{i}+75.0\widehat{j}\right)\text{m}\right)\\ =\left(-227\widehat{i}+79.0\widehat{j}\right)\text{m}\end{array}$

Thus, the coordinates of the particle is $\underset{\_}{\left(-227\text{m , }79.0\text{m}\right)}$.