Chapter 3, Problem 8P

Solution

To Determine:

1. The components of the velocity vector in the north and west directions.
2. The distance in far north and far west covered by the plane after 1.75 h.

Solution:

1. The component of the velocity vector in the northerly direction is 625 km/h and the component of the velocity vector in the westerly direction is -553 km/h.
2. After 1.75 h the airplane travels 1094 km inthe north and 968 km inthe west.

Section 1:

c. To Determine:The components of the velocity vector in the north and west directions.

Solution:The component of the velocity vector in the northerly direction is 625 km/h and the component of the velocity vector in the westerly direction is -553 km/h.

Explanation:

Given: The magnitude of the velocity vector $\left|\overrightarrow{V}\right|$ is 835 km/h and the directionis $41.5˚$ west of north.

Formula used:

The component in the westerly direction is equivalent to x-component, so:

  $V_x=\left|\overrightarrow{V}\right|\cos \theta \mathrm{...}\left(1\right)$

Similarly, the component in the northerly direction is equivalent to y-component, therefore:

  $V_y=\left|\overrightarrow{V}\right|\sin \theta \mathrm{...}\left(2\right)$

Calculation:

Inserting the magnitude of velocity vector and direction into equations (1) and (2) we obtain:

  $\begin{array}{l}{V}_x=\left|\overrightarrow{V}\right|\cos \theta \\ V_x=835\cos \left({90}^o+{41.5}^o\right)\\ V_x=-625\text{km/h}\end{array}$

And

  $\begin{array}{l}{V}_y=\left|\overrightarrow{V}\right|\sin \theta \\ V_y=835\sin \left({90}^o+{41.5}^o\right)\\ V_y=625\text{km/h}\end{array}$

Therefore, the component of the velocity vector in the northerly direction is 625km/h and the component of the velocity vector in the westerly direction is -553 km/h.

Section 2:

1. To Determine:The distance in far north and far west covered by the plane after 1.75 h.

Solution:After 1.75 h the airplane travels 1094 km at the north and 968 km at the west

Explanation:

Given:

The component of the velocity vector in the northerly direction is 625 km/h and the component of the velocity vector in the westerly direction is -553 km/h. The timet is 1.75 h.

Formula used:

Assuming the airplane travels at a constant speed, the distance x traveled in each direction can be obtained using:

  $x=vt$

Where:

• v is the component of vthe elocity vector
• t is the time elapsed

Calculation:

The distance at the north $x_{north}$ is:

  $\begin{array}{c}{x}_{north}=v_{y}t\\ =\left(625\right)\left(1.75\right)\\ =1094\text{km}\end{array}$

Now, the distance at the west $x_{west}$ is:

  $\begin{array}{c}{x}_{west}=v_{x}t\\ =\left(553\right)\left(1.75\right)\\ =968\text{km}\end{array}$

Therefore, after 1.75 h the airplane travels 1094 km at the north and 968 km at the west.

Conclusion:

The component of the velocity vector in the northerly direction is 625 km/h, and the component of the velocity vector in the westerly direction is -553 km/h. After 1.75 h the airplane travels 1094 km at the north and 968 km at the west.ss