Answer to Question #170531 in Mechanics | Relativity for chloe

Question

A yacht is going 10kmh-1 20° north of east relative to the ground when a person slipped on a banana peel and fell off the yacht. He was carried off with the current at 8kmh-1 west. What is the relative velocity of the person in respect to the yacht?

Answer 1

Explanations & Calculations

• Let's write the data that is available in the question

$\qquad\qquad \begin{aligned} \small V_{y,e}&=\small 10kmh^{-1}[\bold{20^0N\,of\,E}]\\ \small V_{m,e}&=\small 8kmh^{-1}[\bold{W}] \end{aligned}$

• All the given measurements are relative to the ground(e)
• What is needed is the velocity of the man relative to the yacht: the motion of man seen by a person on the yacht, $\small V_{m,y}$
• By the equation of relative velocity,

$\qquad\qquad \begin{aligned} \small V_{m,y}&=\small V_{m,e}+V_{e,y}\\ &=\small \overleftarrow{8}[W]+10[20^0\,S\,of\,W] \end{aligned}$

• From here onwards, you can find the resultant velocity vector by creating a vector triangle to some scale or by resolution of vectors on orthogonal directions & get the resultant.
• Therefore,

$\qquad\qquad \begin{aligned} \small \vec{V_{m,y}}&=\small \overleftarrow{8+10\cos20}+\downarrow10\sin20\\ &=\small \overleftarrow{17.397}+\downarrow3.420\\ \small|\vec{V_{m,y}}|&=\small \sqrt{17.397^2+3.420^2}\\ &=\small \bold{17.73\,kmh^{-1}} \end{aligned}$

• Direction of it is

$\qquad\qquad \begin{aligned} \small \theta&= \small \tan^{-1}\Big(\frac{3.420}{17.397}\Big)\\ &=\small \bold{11.12^0 [S\,of\,W]} \end{aligned}$

• Velocity of man relative to yacht is $\small\bold{ 17.73\,kmh^{-1}[11.12^0 \,S\,of\,W]}$