Chapter 4, Problem 5P

(a)

To determine

The components of the particle’s velocity at $t=10.0\sec$.

Answer to Problem 5P

The $x$ components of the particle’s velocity at $t=10.0\sec$ is $-45\text{m}/\text{s}$ and the $y$ component of particle’s velocity at $t=10\text{s}$ is $15\text{m}/\text{s}$.

Explanation of Solution

Formula to calculate net force acts on a particle is,

    ${\overrightarrow{\text{F}}}_{\text{net}}={\overrightarrow{\text{F}}}_1+{\overrightarrow{\text{F}}}_2$

Here, ${\overrightarrow{\text{F}}}_{\text{net}}$ is the net force acting on a particle, ${\overrightarrow{\text{F}}}_1$ and ${\overrightarrow{\text{F}}}_2$ are two forces on the particle.

Formula to calculate acceleration of a particle is,

    $\overrightarrow{a}=\frac{{\overrightarrow{\text{F}}}_{\text{net}}}{m}$

Here, $\overrightarrow{a}$ is the acceleration of the particle and $m$ is the mass of the particle.

Substitute ${\overrightarrow{\text{F}}}_1+{\overrightarrow{\text{F}}}_2$ for ${\overrightarrow{\text{F}}}_{\text{net}}$ in above equation.

    $\overrightarrow{a}=\frac{{\overrightarrow{\text{F}}}_1+{\overrightarrow{\text{F}}}_2}{m}$

Substitute $\left(-6.00\widehat{\text{i}}-4.00\widehat{\text{j}}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_{\text{1}}$, $\left(-3.00\widehat{\text{i}}+7.00\widehat{\text{j}}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_2$ and $2.00\text{kg}$ for $m$ to find $\overrightarrow{a}$.

    $\begin{array}{c}\overrightarrow{a}=\frac{\left(-6.00\widehat{\text{i}}-4.00\widehat{\text{j}}\right)\text{N}+\left(-3.00\widehat{\text{i}}+7.00\widehat{\text{j}}\right)\text{N}}{2.00\text{kg}}\\ =\left(-4.50{\text{m/s}}^{\text{2}}\right)\widehat{\text{i}}+\left(1.50{\text{m/s}}^{\text{2}}\right)\widehat{\text{j}}\end{array}$

Formula to calculate the velocity of the particle is,

    ${\overrightarrow{\text{v}}}_{\text{f}}={\overrightarrow{\text{v}}}_{\text{i}}+\overrightarrow{a}t$

Here, ${\overrightarrow{\text{v}}}_{\text{f}}$ is the final velocity of the particle, ${\overrightarrow{\text{v}}}_{\text{i}}$ is the initial velocity, and $t$ is time.

Substitute $0\text{m}/\text{s}$ for ${\overrightarrow{\text{v}}}_{\text{i}}$, $\left(-4.5\widehat{\text{i}}+1.5\widehat{\text{j}}\right)\text{m}/{\text{s}}^{\text{2}}$ for $\overrightarrow{a}$ and $10.0\text{s}$ for $t$ to find ${\overrightarrow{\text{v}}}_{\text{f}}$.

    $\begin{array}{c}{\overrightarrow{\text{v}}}_{\text{f}}=0+\left[\left(-4.5\widehat{\text{i}}+1.5\widehat{\text{j}}\right)\text{m}/{\text{s}}^{\text{2}}\times \left(10.0\text{s}\right)\right]\\ =\left(-45\widehat{\text{i}}+15\widehat{\text{j}}\right)\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the $x$ components of the particle’s velocity at $t=10.0\sec$ is $-45\text{m}/\text{s}$ and the $y$ component of particle’s velocity at $t=10\text{s}$ is $15\text{m}/\text{s}$.

(b)

To determine

The direction of the motion of the particle at $t=10.0\sec$.

Answer to Problem 5P

The direction of the particle’s motion at $t=10.0\sec$ is $162°$ counterclockwise from the $+x$ axis.

Explanation of Solution

Formula to calculate the direction of the moving particle is,

    $\theta ={\tan }^{-1}\left(\frac{v_{\text{y}}}{v_{\text{x}}}\right)$

Here, $\theta$ is the direction of the velocity of the particle, $v_x$ and $v_y$ are horizontal and vertical components of the particle’s velocity.

Substitute $15\text{m}/\text{s}$ for $v_{\text{y}}$ and $-45\text{m}/\text{s}$ for $v_{\text{x}}$ to calculate $\theta$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{15\text{m}/\text{s}}{-45\text{m}/\text{s}}\right)\\ \theta \approx 162°\end{array}$

Conclusion:

Therefore, the direction of particle’s motion at $t=10.0\sec$ is $162°$ counterclockwise from the $+x$ axis.

(c)

To determine

The displacement of the particle during $10\sec$.

Answer to Problem 5P

The displacement of the particle during $10\sec$ is $\left(-225\widehat{\text{i}}+75\widehat{\text{j}}\right)\text{m}$.

Explanation of Solution

Formula to calculate the displacement of the particle is,

    $\overrightarrow{\text{s}}={\overrightarrow{\text{v}}}_{\text{i}}t+\frac{1}{2}\overrightarrow{a}{t}^2$

Here, $\overrightarrow{\text{s}}$ is the displacement of the particle.

Substitute $0\text{m}/\text{s}$ for ${\overrightarrow{\text{v}}}_{\text{i}}$, $\left(-4.5\widehat{\text{i}}+1.5\widehat{\text{j}}\right)\text{m}/{\text{s}}^{\text{2}}$ for $\overrightarrow{a}$ and $10.0\text{s}$ for $t$ to find $\overrightarrow{\text{s}}$.

    $\begin{array}{c}\overrightarrow{\text{s}}=\left[0\times \left(10.0\text{s}\right)\right]+\left[\frac{1}{2}\left(-4.5\widehat{i}+1.5\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}{\left(10.0\text{s}\right)}^2\right]\\ =\left(-255\widehat{\text{i}}+75\widehat{\text{j}}\right)\text{m}\end{array}$

Conclusion:

Therefore, the displacement of the particle during $10\sec$ is $\left(-255\widehat{\text{i}}+75\widehat{\text{j}}\right)\text{m}$.

(d)

To determine

The coordinates of the particle at $t=10.0\sec$.

Answer to Problem 5P

The coordinates of the particle at $t=10.0\sec$ is $\left(-227\widehat{i}+79\widehat{j}\right)\text{m}$.

Explanation of Solution

The initial position of the particle is $\left(-2\widehat{\text{i}}+4\widehat{\text{j}}\right)\text{m}$.

The final coordinates of the particle at $t=10.0\text{s}$ is,

    $\begin{array}{c}\text{Final coordinates}=\left(-255\widehat{\text{i}}+75\widehat{\text{j}}\right)\text{m}+\left(-2\widehat{\text{i}}+4\widehat{\text{j}}\right)\text{m}\\ =\left(-227\widehat{\text{i}}+79\widehat{\text{j}}\right)\text{m}\end{array}$

Conclusion:

Therefore, the coordinates of the particle at $t=10.0\sec$ is $\left(-227\widehat{\text{i}}+79\widehat{\text{j}}\right)\text{m}$