Chapter 4, Problem 41P

(a)

To determine

The acceleration of the object.

Answer to Problem 41P

Theacceleration of the object is $\left(1.5\text{m}/{\text{s}}^2\right)\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)\widehat{j}$ .

Explanation of Solution

Given:

The mass of the object is $m=4.0\text{kg}$ .

The value of the force ${\overline{F}}_1=\left(2.0\text{N}\right)\widehat{i}+\left(-30\text{N}\right)\widehat{j}$ .

The value of the force ${\overline{F}}_2=\left(4.0\text{N}\right)\widehat{i}-\left(11\text{N}\right)\widehat{j}$ .

Formula used:

The expression for the effective acceleration of the object is given by,

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

Calculation:

The effective acceleration of the object is calculated as,

  $\begin{array}{c}\overrightarrow{a}=\frac{{\overline{F}}_1+{\overline{F}}_2}{m}\\ =\frac{\left(2.0\text{N}\right)\widehat{i}+\left(-30\text{N}\right)\widehat{j}+\left(4.0\text{N}\right)\widehat{i}-\left(11\text{N}\right)\widehat{j}}{4\text{kg}}\\ =\left(1.5\text{m}/{\text{s}}^2\right)\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)\widehat{j}\end{array}$

Conclusion:

Therefore, the acceleration of the object is $\left(1.5\text{m}/{\text{s}}^2\right)\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)\widehat{j}$ .

(b)

To determine

The velocity of the object at time $t=3.0\text{s}$ .

Answer to Problem 41P

Thevelocity of the object at time $t=3.0\text{s}$ is $\left(4.5\text{m}/\text{s}\right)\widehat{i}-\left(10.5\text{m}/{\text{s}}^2\right)\widehat{j}$ .

Explanation of Solution

Given:

Time, $t=3.0\text{s}$

Formula used:

The expression for velocity of the object is given by,

  $\overrightarrow{v}={\displaystyle \int \overrightarrow{a}dt}$

Calculation:

The velocity of the object is calculated as,

  $\begin{array}{c}\overrightarrow{v}={\displaystyle \int \overrightarrow{a}dt}\\ ={\displaystyle \int \left\{\left(1.5\text{m}/{\text{s}}^2\right)\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)\widehat{j}\right\}dt}\\ =\left(1.5\text{m}/{\text{s}}^2\right)t\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)t\widehat{j}+C_1\end{array}$

The velocity of the object for time $t=0$ is calculated as,

  $\begin{array}{l}\overrightarrow{v}=\left(1.5\text{m}/{\text{s}}^2\right)t\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)t\widehat{j}+C_1\\ \left(0\right)=\left(1.5\text{m}/{\text{s}}^2\right)\left(\left(0\right)\right)\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)\left(0\right)\widehat{j}+C_1\\ C_1=0\end{array}$

The evaluated value of velocity of the object for time $t=0$ is calculated as,

  $\overrightarrow{v}=\left(1.5\text{m}/{\text{s}}^2\right)t\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)t\widehat{j}$

The velocity of the object for time $t=3\text{s}$ is calculated as,

  $\begin{array}{c}\overrightarrow{v}=\left(1.5\text{m}/{\text{s}}^2\right)t\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)t\widehat{j}\\ =\left(1.5\text{m}/{\text{s}}^2\right)\left(3\text{s}\right)\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)\left(3\text{s}\right)\widehat{j}\\ =\left(4.5\text{m}/\text{s}\right)\widehat{i}-\left(10.5\text{m}/{\text{s}}^2\right)\widehat{j}\end{array}$

Conclusion:

Therefore, the velocity of the object at time $t=3.0\text{s}$ is $\left(4.5\text{m}/\text{s}\right)\widehat{i}-\left(10.5\text{m}/{\text{s}}^2\right)\widehat{j}$ .

(c)

To determine

The position of the object at time $t=3\text{s}$ .

Answer to Problem 41P

Theposition of the object is $\left(6.75\text{m}\right)\widehat{i}-\left(15.75\text{m}\right)\widehat{j}$ .

Explanation of Solution

Given:

Time, $t=3.0\text{s}$

Formula used:

The expression for position of the object is given by,

  $\overrightarrow{r}={\displaystyle \int \overrightarrow{v}dt}$

Calculation:

The expression for the position of the object is evaluated as,

  $\begin{array}{c}\overrightarrow{r}={\displaystyle \int \overrightarrow{v}dt}\\ ={\displaystyle \int \left\{\left(4.5\text{m}/\text{s}\right)\widehat{i}-\left(10.5\text{m}/{\text{s}}^2\right)\widehat{j}\right\}dt}\\ =\left(1.5\text{m}/{\text{s}}^2\right)\frac{t^2}{2}\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)\frac{t^2}{2}\widehat{j}\end{array}$

The position of the object for time $t=3.0\text{s}$ is given by,

  $\begin{array}{c}\overrightarrow{r}=\left(1.5\text{m}/{\text{s}}^2\right)\frac{{\left(3.0\text{s}\right)}^2}{2}\widehat{i}-\left(3.5\text{m}/{\text{s}}^2\right)\frac{{\left(3.0\text{s}\right)}^2}{2}\widehat{j}\\ =\left(6.75\text{m}\right)\widehat{i}-\left(15.75\text{m}\right)\widehat{j}\end{array}$

Conclusion:

Therefore, the position of the object is $\left(6.75\text{m}\right)\widehat{i}-\left(15.75\text{m}\right)\widehat{j}$ .