Answer to Question #141514 in Mechanics | Relativity for Ariel

Question

The sides of the triangle ABC represents 3 velocities AC(magnitude 3.6m/s) corresponds to the velocity of a boat as observed by a stationary tourist on a bridge. BC (magnitude 1.5m/s) corresponds to the velocity of a cyclist crossing the bridge, once again as seen by the tourist. Calculate the magnitude of the velocity vector AB, the velocity of the boat relative to the cyclist.

Answer 1

Given,

Veocity $AC=3.6m/s$ ,$BC=1.5m/s,$

As for the tourist standing on the bridge

the traingle formed is right angle triangle in which $\angle B=90^{\circ}$

Using pythagoras theorem in $\Delta ABC$

$\Rightarrow AC^2=AB^2+BC^2\\\Rightarrow(3.6)^2=AB^2+(1.5)^2\\\Rightarrow AB^2=12.96-2.25=10.71\\\Rightarrow AB=\sqrt{10.71}=3.27m/s$

Hence the magnitude of velocity of $AB$ is $3.27m/s$