Question #249074

Find the Vr of the vectors given below using both the head-to-tail and component methods.

 

An athlete runs 65.0 km SE the proceeds 50.0 km NE before running 45.0 km NW and 60.0 km W.

Scale = 1.0 cm:10 km


1
Expert's answer
2021-10-10T16:02:48-0400

By the component method:


V1S=65cos45°=46 km,V1E=65sin45°=46 km.V2S=50cos45°=35 km,V2E=50sin45°=35 km.V3S=45cos45°=32 km,V3E=45sin45°=32 km.V4S=60cos90°=0 km,V4E=60sin90°=60 km. VrS=V1S+V2S+V3S+V4S=21 km,VrE=V1E+V2E+V3E+V4E=11 km, Vr=VrE2+VrS2=24 km. θ=arctan2111=62° North of West.V_{1S}=65\cos45°=46\text{ km},\\ V_{1E}=65\sin45°=46\text{ km}.\\ V_{2S}=-50\cos45°=-35\text{ km},\\ V_{2E}=50\sin45°=35\text{ km}.\\ V_{3S}=-45\cos45°=-32\text{ km},\\ V_{3E}=-45\sin45°=-32\text{ km}.\\ V_{4S}=60\cos90°=0\text{ km},\\ V_{4E}=-60\sin90°=-60\text{ km}.\\\space\\ V_{rS}=V_{1S}+V_{2S}+V_{3S}+V_{4S}=-21\text{ km},\\ V_{rE}=V_{1E}+V_{2E}+V_{3E}+V_{4E}=-11\text{ km},\\\space\\ V_r=\sqrt{V^2_{rE}+V^2_{rS}}=24\text{ km}.\\\space\\ \theta=\arctan\frac{21}{11}=62°\text{ North of West.}

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