Chapter 4, Problem 14PQ

An aircraft flies at constant altitude (with respect to sea level) over the South Rim of the Grand Canyon (Fig. P3.40, page 83). Consider a coordinate system such that the positive x axis points to the east, and the positive y axis points north. The aircraft’s initial position and velocity are 1350 m at an angle of 145° and 60.0 m/s at an angle of 55.0° where both angles are measured counterclockwise with respect to the positive x axis. The aircraft’s acceleration is 4.0 m/s² at an angle of 195° with respect to the positive x axis. a. What is the velocity of the aircraft after 7.50 s have elapsed? b. What is the position vector of the aircraft after 7.50 s have elapsed?

To determine

The velocity of the aircraft after $7.50\text{ s}$ have elapsed.

Answer to Problem 14PQ

The velocity of the aircraft after $7.50\text{ s}$ have elapsed is $\left(-1.08\widehat{i}+39.6\widehat{j}\right)\text{m}/\text{s}$.

Explanation of Solution

An aircraft flies at constant altitude with initial position of $1350\text{ m}$ at an angle of $145°$ and velocity of $60.0\text{m}/\text{s}$ at an angle of $55.0°$. The acceleration of the aircraft is $4.0\text{m}/{\text{s}}^2$ at an angle of $195°$.

The two-dimensional motion of an object can be described using the components of a vector with magnitude and direction given by $x=A\cos \theta$ and $y=A\sin \theta$.

The position of the aircraft is $\overrightarrow{r_i}=\left(1350\text{ m}\right)\cos 145°\widehat{i}+\left(1350\text{ m}\right)\sin 145°\widehat{j}=\left(-1110\widehat{i}+774\widehat{j}\right)\text{ m}$.

The velocity of the aircraft is $\overrightarrow{v_i}=\left(60.0\text{m}/\text{s}\right)\cos 55.0°\widehat{i}+\left(60.0\text{m}/\text{s}\right)\sin 55.0°\widehat{j}=\left(34.4\widehat{i}+49.1\widehat{j}\right)\text{m}/\text{s}$.

The acceleration of the aircraft is $\overrightarrow{a}=\left(4.90\text{m}/{\text{s}}^2\right)\cos 195°\widehat{i}+\left(4.90\text{m}/{\text{s}}^2\right)\sin 195°\widehat{j}=\left(-4.73\widehat{i}-1.27\widehat{j}\right)\text{m}/{\text{s}}^2$.

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, $\overrightarrow{v_i}$ is the initial velocity, $t$ is time and $a$ is the acceleration.

Conclusion:

Substitute $\left(34.4\widehat{i}+49.1\widehat{j}\right)\text{m}/\text{s}$ for $\overrightarrow{v_i}$, $7.50\text{ s}$ for $t$ and $\left(-4.73\widehat{i}-1.27\widehat{j}\right)\text{m}/{\text{s}}^2$ for $\overrightarrow{a}$ in equation (II) to find the value of $\overrightarrow{v_f}$

$\begin{array}{l}\overrightarrow{v_f}=\left(34.4\widehat{i}+49.1\widehat{j}\right)\text{m}/\text{s}+\left(-4.73\widehat{i}-1.27\widehat{j}\right)\text{m}/{\text{s}}^2\times t\\ \overrightarrow{v_f}=\left[34.4\widehat{i}+49.1\widehat{j}-4.73t\widehat{i}-1.27t\widehat{j}\right]\text{m}/\text{s}\\ \overrightarrow{v_f}=\left[\left(34.4-4.73\times 7.50\right)\widehat{i}+\left(49.1-1.27\times 7.50\right)\widehat{j}\right]\text{m}/\text{s}\\ \overrightarrow{v_f}=\left(-1.08\widehat{i}+39.6\widehat{j}\right)\text{m}/\text{s}\end{array}$

Thus, the velocity of the aircraft after $7.50\text{ s}$ have elapsed is $\left(-1.08\widehat{i}+39.6\widehat{j}\right)\text{m}/\text{s}$.

To determine

The position of the aircraft after $7.50\text{ s}$ have elapsed.

Answer to Problem 14PQ

The position of the aircraft after $7.50\text{ s}$ have elapsed is $\left(-981\widehat{i}+1110\widehat{j}\right)\text{ m}$.

Explanation of Solution

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 and $\overrightarrow{r_i}$ is the initial position of the object.

Conclusion:

Substitute $\left(-1110\widehat{i}+774\widehat{j}\right)\text{ m}$ for $\overrightarrow{r_i}$, $\left(34.4\widehat{i}+49.1\widehat{j}\right)\text{m}/\text{s}$ for $\overrightarrow{v_i}$, $7.50\text{ s}$ for $t$ and $\left(-4.73\widehat{i}-1.27\widehat{j}\right)\text{m}/{\text{s}}^2$ for $\overrightarrow{a}$  in equation (I) to find the value of $\overrightarrow{r}$

$\begin{array}{l}\overrightarrow{r}=\left(-1110\widehat{i}+774\widehat{j}\right)\text{ m}+\left(34.4\widehat{i}+49.1\widehat{j}\right)\text{m}/\text{s}\times 7.50\text{ s}+\frac{1}{2}\left(-4.73\widehat{i}-1.27\widehat{j}\right){\left(7.50\text{ s}\right)}^2\text{ m}\\ \overrightarrow{r}=\left(\left(-1110+34.4\times 7.50+\frac{1}{2}\left(-4.73\right){\left(7.50\right)}^2\right)\widehat{i}+\left(774+49.1\left(7.50\right)+\frac{1}{2}\left(-1.27\right){\left(7.50\right)}^2\right)\widehat{j}\right)\text{ m}\\ \overrightarrow{r}=\left(-981\widehat{i}+1110\widehat{j}\right)\text{ m}\end{array}$

Thus, The position of the aircraft after $7.50\text{ s}$ have elapsed is $\left(-981\widehat{i}+1110\widehat{j}\right)\text{ m}$.