Chapter 5, Problem 70PQ

A 1.50-kg particle initially at rest and at the origin of an x–y coordinate system is subjected to a time-dependent force of $\stackrel{\to }{F}$(t) = (3.00t $\hat{i}$ − 6.00 $\hat{j}$) N with t in seconds. a. At what time t will the particle’s speed be 15.0 m/s? b. How far from the origin will the particle be when its velocity is 15.0 m/s? c. What is the particle’s total displacement at this time?

To determine

Find the time at which the particle’s speed will be $15.0\text{ m/s}$.

Answer to Problem 70PQ

The time at which the particle’s speed will be $15.0\text{ m/s}$ is $\text{3}\text{.00 s}$.

Explanation of Solution

Write the equation for acceleration.

  $\overrightarrow{a}=\frac{\overrightarrow{F}}{m}$                                                                       (I)

Here, $\overrightarrow{F}$ is the force, $m$ is the mass and $\overrightarrow{a}$ is the acceleration.

Substitute $\left(3.00t\widehat{i}-6.00\widehat{j}\right)\text{ N}$ for $\overrightarrow{F}$ and $1.50\text{ kg}$ for $m$ in equation I.

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

To find the instantaneous velocity integrate the above equation.

    $\begin{array}{c}d\overrightarrow{v}=\left(2.00{\text{ m/s}}^3\right)tdt\widehat{i}-\left(4.00{\text{ m/s}}^2\right)dt\widehat{j}\\ {\displaystyle \int d\overrightarrow{v}}={\displaystyle \int \left(2.00{\text{ m/s}}^3\right)tdt\widehat{i}}-{\displaystyle \int \left(4.00{\text{ m/s}}^2\right)dt\widehat{j}}\\ \overrightarrow{v}=\left[\left(1.00{\text{ m/s}}^3\right)t^2+c_1\right]\widehat{i}-\left[\left(4.00{\text{ m/s}}^2\right)t+c_2\right]\widehat{j}\end{array}$                                    (II)

In order to find the constant of integration consider the object is at rest and $t=0\text{ s}$, then above equation becomes.

    $\overrightarrow{v}\left(t=0\right)=0=\left[\left(1.00{\text{ m/s}}^3\right){\left(0\right)}^2+c_1\right]\widehat{i}-\left[\left(4.00{\text{ m/s}}^2\right)\left(0\right)+c_2\right]\widehat{j}$

Then, $c_1=c_2=0$, then equation II will become,

    $\overrightarrow{v}=\left(1.00{\text{ m/s}}^3\right)t^2\widehat{i}-\left(4.00{\text{ m/s}}^2\right)t\widehat{j}$

Conclusion:

Substitute $\text{15}\text{.0 m/s}$ for $v$ the above equation will become.

    $\left|\overrightarrow{v}\right|=\text{15}\text{.0 m/s}=\sqrt{{\left[\left(1.00{\text{ m/s}}^3\right)t^2\right]}^2-{\left[\left(4.00{\text{ m/s}}^2\right)t\right]}^2}$

  $\begin{array}{l}{\text{225 m}}^2{\text{/s}}^2={\left[\left(1.00{\text{ m/s}}^3\right)t^2\right]}^2-{\left[\left(4.00{\text{ m/s}}^2\right)t\right]}^2\\ \left(1.00{\text{ m}}^2{\text{/s}}^6\right)t^4+\left(4.00{\text{ m}}^2{\text{/s}}^4\right)t^2-{\text{225 m}}^2{\text{/s}}^2=0\end{array}$

The above equation can be factored as,

    $\left(t^2-9\right)\left(t^2+25\right)=0$

Then solution to the above equation to find t is,

    $\begin{array}{c}{t}^2-9.00=0\\ t=\pm 3.00\text{ s}\\ =\text{3}\text{.00 s}\end{array}$

Therefore, the time at which the particle’s speed will be $15.0\text{ m/s}$ is $\text{3}\text{.00 s}$.

To determine

Find the distance at which the particle will travel with velocity $15.0\text{ m/s}$.

Answer to Problem 70PQ

The distance at which the particle will travel with velocity $15.0\text{ m/s}$ is $\text{20}\text{.1 m}$.

Explanation of Solution

In order to find the position of the object the velocity equation should be integrated.

    $\begin{array}{c}d\overrightarrow{r}=\left(1.00{\text{ m/s}}^3\right)t^{2}dt\widehat{i}-\left(4.00{\text{ m/s}}^2\right)tdt\widehat{j}\\ {\displaystyle \int d\overrightarrow{r}}={\displaystyle \int \left(1.00{\text{ m/s}}^3\right)t^{2}dt\widehat{i}}-{\displaystyle \int \left(4.00{\text{ m/s}}^2\right)tdt\widehat{j}}\\ \overrightarrow{r}=\left(0.333{\text{ m/s}}^3\right)t^2\widehat{i}-\left(2.00{\text{ m/s}}^2\right)t^2\widehat{j}\end{array}$                                                 (III)

Conclusion:

Substitute $\text{3}\text{.00 s}$ for $t$ in the above equation.

    $\begin{array}{c}\left|\overrightarrow{r}\right|=\sqrt{{\left[\left(0.333{\text{ m/s}}^3\right){\left(3.00\text{ s}\right)}^2\right]}^2-{\left[\left(2.00{\text{ m/s}}^2\right){\left(3.00\text{ s}\right)}^2\right]}^2}\\ =20.1\text{ m}\end{array}$

Therefore, the distance at which the particle will travel with velocity $15.0\text{ m/s}$ is $\text{20}\text{.1 m}$.

To determine

Find the particle’s total displacement.

Answer to Problem 70PQ

The particle’s total displacement is $\left(9.00\text{ m}\right)\widehat{i}-\left(18.00\text{ m}\right)\widehat{j}$.

Explanation of Solution

Use the equation III to get the total displacement.

Conclusion:

Substitute $\text{3}\text{.00 s}$ for $t$ in equation III.

    $\begin{array}{c}\overrightarrow{r}=\left(0.333{\text{ m/s}}^3\right){\left(3.00\text{ s}\right)}^2\widehat{i}-\left(2.00{\text{ m/s}}^2\right){\left(3.00\text{ s}\right)}^2\widehat{j}\\ =\left(9.00\text{ m}\right)\widehat{i}-\left(18.00\text{ m}\right)\widehat{j}\end{array}$

Therefore, the particle’s total displacement is $\left(9.00\text{ m}\right)\widehat{i}-\left(18.00\text{ m}\right)\widehat{j}$.