Question #141514

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.

Expert's answer

Given,

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


As for the tourist standing on the bridge

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


Using pythagoras theorem in ΔABC\Delta ABC

⇒AC2=AB2+BC2⇒(3.6)2=AB2+(1.5)2⇒AB2=12.96−2.25=10.71⇒AB=10.71=3.27m/s\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 ABAB is 3.27m/s3.27m/s


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