Question

You are riding on a Jet Ski at an angle of 35 degree upstream on a river flowing with a speed of 2.8m/s. If your velocity relative to the ground is 9.5m/s at an angle of 20 degree upstream, what is the speed of the Jet Ski relative to the water. (Note that angles are measured relative to the x axis)

Answer 1

Given information:

The velocity of the river is 2.8 m/s. The jet ski is at angle of 35° upstream. The velocity of the Jet ski relative to the ground is 9.5 m/s at an angle of 20° upstream.

The relative horizontal velocity of the Jet ski relative to the ground is,

`\begin{gathered} \vec{v}_(jgx)=v_(jg)\cos (20\degree)(-\hat{i}) \\ =(9.5\text{ m/s})*\cos (20\degree)(-\hat{i}) \\ \approx(8.927\text{ m/s})(-\hat{i}) \end{gathered}`

The relative vertical velocity of the Jet ski relative to the ground is,

`\begin{gathered} \vec{v_(jgy)}=v_(jg)\sin (20\degree)(\hat{j}) \\ =(9.5\text{ m/s})*\sin (20\degree)(\hat{j}) \\ \approx(3.249\text{ m/s})(\hat{j}) \end{gathered}`

The relative velocity of the jet ski relative to the ground is,

`\begin{gathered} \vec{v}=\vec{v}+\vec{v} \\ =(8.927\text{ m/s})(-\hat{i})+(3.249\text{ m/s})(\hat{j}) \end{gathered}`

The relative velocity of the river relative to the ground is,

`\begin{gathered} \vec{v}_(rg)=v_(rg)(\hat{j}) \\ =(-2.8\text{ m/s})(\hat{j}) \end{gathered}`

The relative velocity of the Jet ski relative to the river water is,

`\begin{gathered} \vec{v}=\vec{v}-\vec{v} \\ =(8.3\text{ m/s})(-i)+(3.249\text{ m/s})(\hat{j})-(-2.8\text{ m/s})(\hat{j}) \\ =(8.927\text{ m/s})(-i)+(6.049\text{ m/s})(\hat{j}) \end{gathered}`

The magnitude of the relative velocity of the Jet ski relative to the river water is,

`\begin{gathered} v_(jr)=\sqrt[]{(8.927\text{ m/s})^2+(6.049\text{ m/s})^2} \\ \approx10.8\text{ m/s} \end{gathered}`

Therefore, the speed of the Jet ski relative to the water is 10.8 m/s.