We have the following excerpt for the analysis:
" A ship is travelling due north with a speed of 12 km/hr relative to the water. The current in the water is flowing at 1.1 km/hr to the east. Determine the magnitude and the velocity of the ship "
We will investigate the application for the concept of relative velocity. The following variables can be extracted from the given excerpt:
![\begin{gathered} V_(sw)\text{ = 12 }\frac{\operatorname{km}}{hr}\ldots\text{ due north} \\ \\ V_w\text{ = }1.1\text{ }\frac{\operatorname{km}}{hr}\ldots\text{ due east} \end{gathered}]()
In such cases of application we seek help by formulating a velocity triangle. We need to determine the direction and magnitude of ship's velocity that it must traverse in and by the effect of current due east the ship starts to move north.
We can go ahead and express the above information graphically in form a velocity triangle as follows:
Think of the above velocity triangle as such:
" The ship initital moves with a velocity ( Vs ) which is unknown at a certain angle. During the course the current of the water ( Vw ) pushes the ship nose from east side towards north. With the effect of the current the ship "seems" to be moving due north ( Vsw ) "
The above triangle is represented as a right angled triangle. We will recall the properties and the application of pythagoream theorem that only applies to right angle triangle as follows:
Where,
We will plug in the respective quantities in the pythagorean theorem relation expressed above:
![\begin{gathered} (V_s)\text{ = }\sqrt[]{(V_w)^2+(V_(sw))^2} \\ \\ (V_s)\text{ = }\sqrt[]{(1.1)^2+(12_{})^2} \\ \\ (V_s)\text{ = }\sqrt[]{(145.21)} \\ \\ (V_s)\text{ = 12.05 }\frac{\operatorname{km}}{hr}\ldots Answer \end{gathered}]()