The absolute velocity of Ship B, relative to the Earth, is approximately 9.9 km/h at an angle of approximately 60 degrees north of east.
To solve this problem, we can use vector addition to find the absolute velocity of Ship B. The velocity of B relative to A is a vector, and we need to add it to the velocity of A to find the absolute velocity of B.
Let
be the velocity vector of Ship A, and
be the velocity vector of Ship B relative to A. The absolute velocity of Ship B
is given by the vector sum:
![\[ \mathbf{V}_B = \mathbf{V}_A + \mathbf{V}_(B/A) \]](https://img.qammunity.org/2024/formulas/physics/high-school/knsb13c48shbla7zk8zncauqe7oiqvj43h.png)
First, we need to express the velocities in vector form. The velocity vector \( \mathbf{V}_A \) is given by:
![\[ \mathbf{V}_A = 12 \, \text{km/h} \, \angle \, 30^\circ \]](https://img.qammunity.org/2024/formulas/physics/high-school/h7v923oa95amifw3onq8449vtjvavev39b.png)
The velocity vector
is given by:
![\[ \mathbf{V}_(B/A) = 10 \, \text{km/h} \, \angle \, 270^\circ \]](https://img.qammunity.org/2024/formulas/physics/high-school/u7a8s7op4xc1d0py4gd14wb2sw8y0fjeud.png)
Now, let's perform the vector addition:
![\[ \mathbf{V}_B = \mathbf{V}_A + \mathbf{V}_(B/A) \]](https://img.qammunity.org/2024/formulas/physics/high-school/knsb13c48shbla7zk8zncauqe7oiqvj43h.png)
![\[ \mathbf{V}_B = 12 \, \text{km/h} \, \angle \, 30^\circ + 10 \, \text{km/h} \, \angle \, 270^\circ \]](https://img.qammunity.org/2024/formulas/physics/high-school/74bf42kg2w7e7ke6019bcjqczysg3td94u.png)
To simplify this, we need to convert the vectors to Cartesian coordinates:
![\[ \mathbf{V}_A = 12 \cos(30^\circ) \, \hat{i} + 12 \sin(30^\circ) \, \hat{j} \]](https://img.qammunity.org/2024/formulas/physics/high-school/1huounkf1hsgqq4upk1y3sgcahiuqihm9r.png)
![\[ \mathbf{V}_(B/A) = 10 \cos(270^\circ) \, \hat{i} + 10 \sin(270^\circ) \, \hat{j} \]](https://img.qammunity.org/2024/formulas/physics/high-school/jmwce9o8mprewhpe11yindsyyhdome7n8m.png)
Performing the vector addition:
![\[ \mathbf{V}_B = (12 \cos(30^\circ) + 10 \cos(270^\circ)) \, \hat{i} + (12 \sin(30^\circ) + 10 \sin(270^\circ)) \, \hat{j} \]](https://img.qammunity.org/2024/formulas/physics/high-school/uvhxk8cd4z8hk8jdc7k2fbzkiqtvo1nt2v.png)
Now, calculate the magnitudes and angles:
![\[ \text{Magnitude of } \mathbf{V}_B = \sqrt{(\text{coefficient of } \hat{i})^2 + (\text{coefficient of } \hat{j})^2} \]](https://img.qammunity.org/2024/formulas/physics/high-school/f029xf45p64vn0iwvhb5jr6ime3anx4cga.png)
![\[ \text{Direction of } \mathbf{V}_B = \arctan\left(\frac{\text{coefficient of } \hat{j}}{\text{coefficient of } \hat{i}}\right) \]](https://img.qammunity.org/2024/formulas/physics/high-school/f4uuvc0hkz00x3u2nx2r3oo56lv1vid79o.png)
The probable question maybe:
If Ship A is moving at a steady speed of 12 km/h in a direction of 30 degrees, and Ship B has a relative velocity of 10 km/h in a direction of 270 degrees with respect to Ship A, what is the magnitude and direction of Ship B's velocity in an absolute reference frame?