Chapter 4, Problem 11PQ

(a)

To determine

Write the position vector of the object as a function of time.

Answer to Problem 11PQ

The position vector of the object as a function of time is $\left(1.25t^2\widehat{i}+3.00t\widehat{j}\right)\text{ m}$.

Explanation of Solution

An object is moving with a initial velocity $\overrightarrow{v_i}=3.00\widehat{j}\text{m}/\text{s}$ and with acceleration $\overrightarrow{a}=2.50\widehat{i}\text{m}/{\text{s}}^{\text{2}}$ and the object was initially at the origin $(0,0)$.

Write the formula for the position vector [two-dimensional kinematic equation]

    $\overrightarrow{r}=\overrightarrow{r_i}+\overrightarrow{v_i}t+\frac{1}{2}a{t}^2$ (I)

Here, $\overrightarrow{r}$ is the final position vector, $\overrightarrow{r_i}$ is the initial position of the object, $\overrightarrow{v_i}$ is the initial velocity, $t$ is time and $a$ is the acceleration.

Conclusion:

The initial position vector is $\overrightarrow{r_i}=0\widehat{i}+0\widehat{j}$ as the object was initially at the origin.

Substitute $3.00\widehat{j}\text{m}/\text{s}$ for $\overrightarrow{v_i}$, $0\widehat{i}+0\widehat{j}$ for $\overrightarrow{r_i}$ and $2.50\widehat{i}\text{m}/{\text{s}}^{\text{2}}$ for $\overrightarrow{a}$ in equation (I) to find the value of $\overrightarrow{r}$

$\begin{array}{l}\overrightarrow{r}=0\widehat{i}+0\widehat{j}+3.00\widehat{j}\text{m}/\text{s}\times t+\frac{1}{2}\left(2.50\widehat{i}\text{m}/{\text{s}}^{\text{2}}\right)t^2\\ \overrightarrow{r}=\left(1.25t^2\widehat{i}+3.00t\widehat{j}\right)\text{ m}\end{array}$

Thus, the position vector of the object as a function of time is $\left(1.25t^2\widehat{i}+3.00t\widehat{j}\right)\text{ m}$.

(b)

To determine

Write the velocity vector of the object as a function of time.

Answer to Problem 11PQ

The velocity vector of the object as a function of time is $\left(2.50t\widehat{i}+3.00\widehat{j}\right)\text{m}/\text{s}$.

Explanation of Solution

Write the formula for the velocity vector [two-dimensional kinematic equation]

    $\overrightarrow{v_f}=\overrightarrow{v_i}+at$ (II)

Here, $\overrightarrow{v_f}$ is the final velocity vector.

Conclusion:

Substitute $3.00\widehat{j}\text{m}/\text{s}$ for $\overrightarrow{v_i}$ and $2.50\widehat{i}\text{m}/{\text{s}}^{\text{2}}$ for $\overrightarrow{a}$ in equation (II) to find the value of $\overrightarrow{v_f}$

$\begin{array}{l}\overrightarrow{v_f}=3.00\widehat{j}\text{m}/\text{s}+2.50\widehat{i}\text{m}/{\text{s}}^{\text{2}}t\\ \overrightarrow{v_f}=\left(2.50t\widehat{i}+3.00\widehat{j}\right)\text{m}/\text{s}\end{array}$

Thus, the velocity vector of the object as a function of time is $\left(2.50t\widehat{i}+3.00\widehat{j}\right)\text{m}/\text{s}$.

(c)

To determine

Write the position vector of the object at time $t=3.00\text{ s}$.

Answer to Problem 11PQ

The position vector of the object at time $t=3.00\text{ s}$ is $\left(11.3\widehat{i}+9.00\widehat{j}\right)\text{ m}$.

Explanation of Solution

The position of the object after $3.00\text{ s}$ is

Substitute $3.00\widehat{j}\text{m}/\text{s}$ for $\overrightarrow{v_i}$, $0\widehat{i}+0\widehat{j}$ for $\overrightarrow{r_i}$, $3.00\text{ s}$ for $t$ and $2.50\widehat{i}\text{m}/{\text{s}}^{\text{2}}$ for $\overrightarrow{a}$ in equation (I) to find the value of $\overrightarrow{r}$

$\begin{array}{l}\overrightarrow{r}=0\widehat{i}+0\widehat{j}+3.00\widehat{j}\text{m}/\text{s}\times 3.00\text{ s}+\frac{1}{2}\left(2.50\widehat{i}\text{m}/{\text{s}}^{\text{2}}\right){\left(3.00\text{ s}\right)}^2\\ \overrightarrow{r}=\left(1.25\times {\left(3.00\right)}^2\widehat{i}+3.00\times 3.00\widehat{j}\right)\text{ m}\\ \overrightarrow{r}=\left(11.3\widehat{i}+9.00\widehat{j}\right)\text{ m}\end{array}$

Thus, the position vector of the object at time $t=3.00\text{ s}$ is $\left(11.3\widehat{i}+9.00\widehat{j}\right)\text{ m}$.

(d)

To determine

Write the speed of the object at time $t=3.00\text{ s}$.

Answer to Problem 11PQ

The speed of the object at time $t=3.00\text{ s}$ is $8.08\text{m}/\text{s}$.

Explanation of Solution

The velocity of the object after $3.00\text{ s}$ is

Substitute $3.00\widehat{j}\text{m}/\text{s}$ for $\overrightarrow{v_i}$, $3.00\text{ s}$ for $t$ and $2.50\widehat{i}\text{m}/{\text{s}}^{\text{2}}$ for $\overrightarrow{a}$ in equation (II) to find the value of $\overrightarrow{v_f}$

$\begin{array}{l}\overrightarrow{v_f}=3.00\widehat{j}\text{m}/\text{s}+2.50\widehat{i}\times 3.00\text{m}/\text{s}\\ \overrightarrow{v_f}=\left(7.50\widehat{i}+3.00\widehat{j}\right)\text{m}/\text{s}\end{array}$

Thus, the velocity vector of the object at time $t=3.00\text{ s}$ is $\left(7.50\widehat{i}+3.00\widehat{j}\right)\text{m}/\text{s}$.

The speed of the object is $v_f=\sqrt{v^2{}_{xf}+v^2{}_{yf}}=\sqrt{{\left(7.50\right)}^2+{\left(3.00\right)}^2}=8.08\text{m}/\text{s}$.

Thus, the speed of the object at time $t=3.00\text{ s}$ is $8.08\text{m}/\text{s}$.